A 10kg box slides 4.0 m down the frictionless ramp shown in the figure, then collides with a spring whose spring constant is 200 N/m.(a) What is the maximum compression of the spring? (b) Now consider a new situation when the box is resting against the spring. No motion. The box is in equilibrium with the spring. What is the compression in the spring?

You attach a 1.90 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by 0.500 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches a speed of zero again 0.500 s after release (for the first time after release). What is the maximum speed of the block (in m/s)?

As shown in the figure, a 1.50-kg box is held at rest against a spring with a force constant k = 790 N/m that is compressed a distance d. When the box is released, it slides across a surface that is frictionless, except for a rough patch that has a coefficient of kinetic friction u, = 0.400 and is 6.00 cm in length. If the speed of the box is 1.50 m/s after sliding across the rough patch, determine the initial compression d of the spring. cm Equilibrium position v=0' rough patch wwwwwn 6.00 cm

A 3.00 kg box that is sliding on frictionless surface with a speed of 15 m/s approaches a horizontal spring. The spring has a spring constant of 1900 N/m. m 10000000 How far will the spring be compressed in stopping the box? Submit Answer Tries 0/10 How far will the spring be compressed when the box's speed is reduced to half of its initial speed? Note: This is not saying that the box starts with half of the original speed, the box still starts with 15 m/s, the question asks how far the spring will have been compressed when it has slowed the box to half of that speed. Submit Answer Tries 0/10

A block sitting on a frictionless surface is attached to a spring of spring constant k. The other end of the spring is attached to the wall. The equilibrium length of the spring is x1. The spring is first compressed to a distance of x2, then released and extended to a length of x3. The elastic potential energy stored in the spring when it is extended to x3 can be calculated as: х, wwWO wwWO A block oscillating on the end of a spring Pick the correct answer PE = }ka? PE = }k(x3 – x2)² PE = 서(z3-z) PE = }k(x3 – x1)² PE = }k(x} – a})

A 25.0-kg block is released from rest at point A. The track is frictionless except for the portion between points B and C, which has a length of 10.00 m. The block travels down the track, hits a spring of force constant 2000 N/m, and compresses the spring 0.50 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points B and C.

A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determine the force constant (in N/m) of the spring, if the box compresses the spring 4.7 cm before coming to rest. N/m (b) Determine the initial speed (in m/s) the box would need in order to compress the spring by 1.5 cm. m/s Additional Materials O Reading

A 500-g block is released from rest and slides down a frictionless track that begins 1.86 m above the horizontal, as shown in the figure below. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring constant of 16.0 N/m. Find the maximum distance the spring is compressed. 50 0g h k

Initially sliding with a speed of 4.3 m/sm/s, a 2.0 kg block collides with a spring and compresses it 0.42 mm before coming to rest.What is the force constant of the spring?

One end of a horizontal spring with k=80N/m is fixed on the wall. A 5kg block is moving at 2m/s when it touches on the free end of a spring. Find the maximum compression of spring Δ: a) if the floor is frictionless ... Δa=...b) if there is kinetic friction with magnitude 10N between block and the floor..... Δb=...c) the amount of thermal energy produced in case (a).. ΔEth-a=.... and (b)... ΔEth-b=.......