214k views
5 votes
A 1. 80-kg block slides on a rough, level surface. When it is traveling at 2. 00 m/s, the block hits a linear, ideal spring and compresses it by a distance of 11 cm before (momentarily) coming to rest. The coefficient of kinetic friction between the block and the surface is 0. 56. What is the force constant of the spring?.

User Bhell
by
3.2k points

1 Answer

3 votes

Use the work-energy theorem.

The total work W performed on the block is done by friction and the restoring force in the spring, both of which oppose the block's motion. By Newton's second law,

• the net horizontal force on the block is

∑ F (hor) = -f - s = ma

where f = magnitude of friction, s = mag. of restoring force, m = 1.80 kg, and a is the acceleration of the block; and

• the net vertical force on the block is

∑ F (ver) = n - mg = 0

where n = mag. of normal force.

It follows that

n = mg ⇒ f = 0.56n = 0.56mg

and so the work done by friction as the spring is compressed by 11 cm = 0.110 m is

W (friction) = -0.56mg (0.110 m) = -1.09 J

The work done by the spring as it compresses is

W (restoring) = -1/2 k (0.110 m)²

where k is the spring constant.

Then the total work is

W = -1.09 J - 1/2 k (0.110 m)²

By the work-energy theorem, this work is equal to the change in the block's kinetic energy,

W = ∆K

Since it comes to a stop, we have

∆K = 0 - 1/2 mv² = -1/2 (1.80 kg) (2.00 m/s)² = -3.60 J

So, we have

-1.09 J - 1/2 k (0.110 m)² = -3.60 J

and solving for k gives

k = (-3.60 J + 1.09 J) / (-1/2 (0.110 m)²)

k ≈ 415 N/m

User Flooooo
by
3.2k points