Question

The spring shown in the Figure below is compressed 50 cm and used to launch a 100 kg physics student. The track is frictionless until it starts up the incline. The student’s coefficient of kinetic friction on the 30◦ incline is 0.15. (a) What is the student’s speed just after losing contact with the spring?

(b) What is the student’s speed immediately before moving up the 30◦ incline?
(c) How far up the incline does the student go?

Answer 1

The potential energy stored in the block-spring-support system when the block is just released is 1.25 J. The speed of the block when it crosses the point where the spring is neither compressed nor stretched is approximately √2.5 m/s. The block comes to rest approximately 25.5 cm up the incline.

Step-by-step explanation:

To answer part (a), we can use the principle of conservation of mechanical energy. The potential energy stored in the block-spring-support system can be calculated using the formula PE = 0.5kx^2, where k is the force constant of the spring and x is the compression of the spring. In this case, the compression is 10 cm or 0.1 m. Plugging in the values, we get PE = 0.5(250 N/m)(0.1 m)^2 = 1.25 J. Therefore, 1.25 J of potential energy is stored in the block-spring-support system when the block is just released.

To answer part (b), we can equate the potential energy stored in the block-spring-support system to the kinetic energy of the block when it crosses the point where the spring is neither compressed nor stretched. The formula for kinetic energy is KE = 0.5mv^2, where m is the mass of the block and v is its velocity. Setting PE equal to KE, we have 1.25 J = 0.5(0.1 kg)v^2. Solving for v, we find v = √2.5 m/s. Therefore, the speed of the block when it crosses the point where the spring is neither compressed nor stretched is √2.5 m/s.

Part (c) asks us to determine the position of the block where it just comes to rest on its way up the incline. At this point, all the mechanical energy of the block has been converted into potential energy, as it is momentarily at rest. Therefore, we can equate the potential energy at this point to the kinetic energy at the point where the spring is neither compressed nor stretched. Using the formula PE = mgh, where h is the height and g is the acceleration due to gravity, we have mgh = 0.5(0.1 kg)v^2. Plugging in the values, we get (0.1 kg)(9.8 m/s^2)h = 0.5(0.1 kg)(√2.5 m/s)^2. Solving for h, we find h = 0.255 m. Therefore, the block comes to rest approximately 25.5 cm up the incline.