Physics Problem, help needed

Question

A 1.4-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spring is attached to a wall, as shown. The initial height of the block is 0.32 m above the lowest part of the slide and the spring constant is 409 N/m.

(a) What is the block's speed when it is at a height of 0.25 m above the base of the slide? _______ m/s

(b) How far is the spring compressed? ____ m

(c) The spring sends the block back to the left. How high does the block rise? ___m

Answer 1

Problems like this are best solved using the conservation of energy. Conservation of energy means that at any point in time, the total energy of the system will be the same as at any other moment in time. Total energy, in this problem, consists of three different types, gravitational potential energy (PE_gravity), kinetic energy (KE), and spring potential energy (PE_spring). Here are the equations that define each of these:

PE_gravity = mgh

KE = 1/2mv²

PE_spring = 1/2kx²

When we place a 1.4-kg block at a height of 0.32 m above the lowest part of the slide, we give it a gravitational potential energy calculated by PE_gravity = mgh = (1.4)(9.8)(0.32) = 4.3904 joules. It has no kinetic energy at this point because it has no velocity. At all times in its path of motion, it will have 4.3904 joules although sometimes it will have less potential energy and more kinetic energy, but the total will always be 4.3904 joules.

At a height of 0.25 m above the base of the slide, the gravitational potential energy is defined by PE_gravity = mgh = (1.4)(9.8)(0.25) = 3.43 joules. That leaves 4.3904 - 3.43 or 0.9604 joules as kinetic energy. If KE = 1/2mv² and KE = 0.9604 joules then we can say 0.9604 = 1/2(1.4)v² or v = √[2*0.9604)/1.4] = 1.2 m/s

When the block completely compresses the spring, since the velocity will again be zero and since the height above the base is also zero, then all of the 4.3904 joules will have been converted to spring potential energy defined by PE_spring = 1/2kx² so we can say 4.3904 = 1/2(409)x² or x = √[2*4.3904)/409] = 0.15 m

When the block shoots out of the spring, it will again always have 4.3904 joule anywhere along its path of motion. So, when it gets to its peak height, again because the velocity will be zero, it will have no kinetic energy and all gravitational potential energy as defined by PE_gravity = mgh so we can say 4.3904 = (1.4)(9.8)(h) and h will equal (again) 0.32 m