Question

A 20 kg crate is allowed to slide down a 60-m long smooth inclined plane which is 30° with the horizontal. At the bottom of the inclined plane, the crate is allowed to slide horizontally with the ground where the coefficient of friction between the crate and the ground is 0.15. After 1 m, the crate would meet a coil with a spring constant of 1000 N/m. (Assume that the friction no longer acts as the crate hits the spring).Calculate the following The potential energy of the crate when it is on the top The kinetic energy when it is on the base of the incline of the incline. and its speed at this point. The frictional force between the ground and the crate. The work done by the friction in slowing down the crate The Kinetic Energy of the crate just before it hits the The velocity of the crate as it hits the spring spring The maximum Elastic Potential Energy that can be absorbed by the spring. The deformation x that is done by the crate to the spring. Would the crate go up back to the incline? If yes up to what height? If no, where in the ground would it stop.

Answer 1

1. It is a 60m long incline at an angle of 30 degrees, Hence the initial height of the crate is,

$h = 60 sin(30^\circ) = 30 \, m$

Therefore the potential energy is,

$E = mgh \Rightarrow E = (20)(9.81)(30) = 5886 \, J$

2. Once the crate comes down from the incline, the Kinetic energy will be equivalent to the initial potential energy. This gives,

$KE= E \Rightarrow KE = 5886 \, J$

$\Rightarrow \frac{1}{2}mv^2 = 5886 \, J$

$\Rightarrow \frac{1}{2}(20)v^2 = 5886 \, J$

$\Rightarrow v = \sqrt{588.6} = 24.26 \, m/s$

3. The frictional force is given by,

$F_r = \mu N$

Where the normal reaction N is given by the weight of the body. Hence,

$\Rightarrow F_r = \mu mg$

Substituting values we get,

$\Rightarrow F_r = (0.15)(20)(9.81) = 29.43 \, N$

4. The friction acts on the spring for a distance of 1 m, Hence the amount of work done is, (The frictional force is negative because it is opposite to the direction of displacement).

$W = F_r x \Rightarrow W = (-29.43)(1) = -29.43 \, J$