 admin # A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slides a distance of 2.7 m in 5.8 s. Find the coefficient of kinetic friction between the block and plane.

2 months ago

## Solution 1 Guest #9800656
2 months ago

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

## 📚 Related Questions

Question
An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g 10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?
Solution 1

The work by tension is ¹¹/₁₀ Mg

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### Further explanation

Complete Question:

An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g/10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?

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Given:

Mass of the object = M

Acceleration of the object = g/10

Distance =

Work by Tension = W = ?

Solution:

Let's find the magnitude of tension as follows:

$$\Sigma F = ma$$

$$T - Mg = Ma$$

$$T = Mg + Ma$$

$$T = M(g + a)$$

$$T = M(g + \frac{1}{10}g)$$

$$T = M(\frac{11}{10}g)$$

$$T = \frac{11}{10}Mg$$

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$$W = T \times L$$

$$W = \frac{11}{10}Mg \times L$$

$$W = \frac{11}{10}MgL$$

$$\texttt{ }$$

The work by tension is ¹¹/₁₀ Mg

$$\texttt{ }$$

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Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Solution 2

The work done by tension in string is $$\dfrac{11}{10}MgL}$$.

Explanation:

Given data:

Mass of object is, M.

Acceleration of object is, a = g/10.

Distance covered vertically is, L.

The work done by tension in the string is given as,

$$W = T \times L$$  .......................................................... (1)

Here, T is the tension force on string.

Apply the equilibrium of forces on string as,

$$T- Mg=Ma$$

Here, g is gravitational acceleration.

$$T- Mg=M(\dfrac{g}{10} )\\T=M(\dfrac{g}{10} )+Mg\\T=\dfrac{11}{10}Mg$$

Substituting value in equation (1) as,

$$W = \dfrac{11}{10}Mg \times L\\W = \dfrac{11}{10}MgL}$$

Thus, the work done by tension in string is $$\dfrac{11}{10}MgL}$$.

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Question
A tank holds a 1.44-m thick layer of oil that floats on a 0.98-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.40 and 1.52, respectively. A ray originating at O crosses the brine-oil interface at a point 0.7 m from the axis. The ray continues and emerges into the air above the oil. What is the angle that the ray in the air makes with the vertical axis
Solution 1

Angle of ray makes with the vertical is 62.1 degree

Explanation:

As per the ray diagram we know that the angle of incidence on oil brine interface will be given as

$$tan\theta_i = \frac{0.7}{0.98}$$

$$\theta_i = 35.5^0$$

now by Snell'a law at that interface we have

$$\mu_1 sin\theta_i = \mu_2 sin \theta_r$$

now we will have

$$1.52 sin35.5 = 1.40 sin\theta_r$$

$$\theta_r = 39.12^0$$

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

$$\mu_2 sin\theta_i' = \mu_{air} sin\theta$$

$$1.40 sin39.12 = 1 sin\theta$$

$$\theta = 62.1^0$$

Question
A rope attached to a load of 175 kg bricks Ilifts the bricks with a steady acceleration of 0.12.m/s^2 straight up. What is the tension in the rope? (a)2028N (b)1645 N (c) 1894 N (d) 1976 N (e) 1736 N (f) 1792 N
Solution 1

Tension, T = 1736 N

Explanation:

It is given that,

Mass of bricks, m = 175 kg

A rope is attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12 m/s² in vertically upwards direction. let T is the tension in the rope. Using second equation of motion as :

T - mg = ma

T = ma + mg

T = m(a + g)

T = 175 kg ( 0.12 m/s² + 9.8 m/s² )

T = 1736 N

Hence, the tension in the wire is 1736 N.

Solution 2

The tension in the rope is 1736 N.

(e) is correct option.

Explanation:

Given that,

Mass of bricks = 175 kg

Acceleration = 0.12 m/s²

Let T is the tension in the rope.

A rope attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12.m/s^2 in vertically upward direction.

Using equation of balance

$$T-mg=ma$$

$$T=mg+ma$$

$$T=175(9.8+0.12)$$

$$T= 1736\ N$$

Hence, The tension in the rope is 1736 N.

Question
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? (k = 1/4πε0 = 8.99 × 109 N · m2/C2)
Solution 1

see attachment

Explanation:

Solution 2

The electric potential varies inversely with the distance. So, the third charge should be placed at a distance of 3 cm from the origin on the x-axis.

### What is electric potential?

The work done on an electric charge to shift it from infinity to a point is known as electric potential at that point. And its expression is,

$$V = \dfrac{kq}{r}$$

here, k is the coulomb's constant.

Given data:

The magnitude of two point charges are, $$+2.0 \;\rm \mu C$$  and  $$-6.0 \;\rm \mu C$$.

The location of each charge on the x-axis is -1.0 cm and +2.0 cm.

Let the third charge ( $$+3.0 \;\rm \mu C$$ ) be placed at a distance of x. Then the electric potential at origin is,

$$V = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}$$

Since, potential at origin is zero (V = 0). Then,

$$0 = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}\\\\\dfrac{k \times 6.0}{0.02} = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times 3.0}{x}\\\\\dfrac{6.0}{0.02} = \dfrac{2.0}{0.01} +\dfrac{3.0}{x}\\\\x = 0.03 \;\rm m =3 \;\rm cm$$

Thus, we can conclude that the third charge should be placed at a distance of 3 cm from the origin on the x-axis.

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Question
what net force is needed to give a 48.0kg grocery cart an acceleration of 3.39m/s^2 ?
Solution 1

you use the formula F=ma to get the answer

Question
A nuclear power plant operates at 40.0% efficiency with a continuous production of 1192 MW of usable power in 1.00 year and consumes 1.22×106 g of uranium-235 in this time period. What is the energy in joules released by the fission of a single uranium-235 atom?
Solution 1

the answer is in the picture below......

Question
Consider two isolated, charged conducting spheres: a large sphere and a second smaller sphere with a radius 6 times smaller than that of the large sphere, but with 3 times as much charge.

(a) Calculate the ratio of the electric potential at the surface of the large sphere to that of the small sphere.
Solution 1

Let the bigger sphere be sphere 1 and the let the smaller sphere be sphere 2. Rest of the answer is in the picture.

Question
Consider a solid sphere of radius R = 0.4 m that is uniformly charged with ? = -11 ?C/m3. What is the electric potential a distance 5 m from the center of the sphere?
Solution 1

V=-5304.6V

Explanation:

(I set the charge density unit to microcoulombs per metre cubed)

In the picture.

Question
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball traveling horizontally at 37.0 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction.)
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball? (Indicate the direction with the sign of your answer.)
Solution 1

(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s

(b) The work that the racket does on the ball is -35.1 Joule

### Further Explanation

Given :

mass of ball = m = 0.06 kg

initial velocity = v₁ = -50.4 m/s

final velocity = v₂ = 37.0 m/s

Unknown :

(a) Impulse = I = ?

(b) Work = W = ?

Solution :

## Question (a) :

In this question , we could use the formula from Second Law of Newton :

$$I = \Delta p$$

$$I = p_2 - p_1$$

$$I = m \times v_2 - m \times v_1$$

$$I = m \times (v_2 - v_1)$$

$$I = 0.06 \times (37.0 - (-50.4))$$

$$I = 0.06 \times (87.4)$$

$$I = 5.244~kg.m/s$$

$$\large { \boxed {I \approx 5.24~kg.m/s} }$$

## Question (b) :

$$W = F \times d$$

$$W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)$$

$$W = \frac{I(v_1 + v_2)}{2}$$

$$W = \frac{5.244(-50.4 + 37)}{2}$$

$$W = \frac{5.244(-13.4)}{2}$$

$$W = -35.1348~Joule$$

$$\large { \boxed {W \approx -35.1~Joule}}$$

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Subject: Physics

Chapter: Dynamics

Keywords: Newton, Law, Impulse, Work

Question
An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. scaffold weighs 500-N and the worker stands 1.0 m from one end, what is the tension in the rope farther from the worker? (a) 1300 N (b) 1800 N (c) 450 N (d) None of these.
Solution 1

This problem is best answered by drawing a figure as a first step.

ABC is the scaffold.

A downward force of 500N is applied downwards at 1m from end A.

The weight of 800N is exerted by the scaffold uniformly distributed between A & C.

At A and C, ropes are attached to support the load.

Let Fc=tension in rope passing through C.

Fc = (500N * 1m +800N*(3+1)/2m / 4m

= (500 Nm + 1600Nm) / 4m

= 2100 Nm / 4m

= 525 N

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