Question

Starting with a speed of 2.50 m/s, a 2.00 kg block slides across a rough horizontal surface until it collides with a spring. The spring compresses a distance of 6.00cm as the block comes momentarily to rest. The total distance traveled by the block from its starting position to the position where it comes to rest is 0.800m. If the coefficient of kinetic friction between the block and the surface is 0.100, what is the spring constant of the spring?

Answer 1

total distance travelled=0.8m

displacement of spring=6cm=0.06m

Balance distance=0.8-0.06=0.74m (horizontal distance   S)

Apply the below equation for S

=+2aS

u=initial velocity=2.5m/s

v=final velocity=?

S=0.74m

a=retardation

Here the retarding force is friction=f=-mg

=coefficient of friction=0.1

m=mass=2kg

g=9.8m/s2

f=-mg

a=f/m=-g=-0.1X9.8=-0.98m/s2

=+2aS

=+2X(-0.98)X0.74=6.25-1.45=4.8

v=2.19m/s=velocity with which it hits the spring

This kinetic energy of body is converted into elastic potential energy of spring

m/2=k/2

k=spring constant=?

x=compression of spring=0.06m

m/2=k/2

2X4.8/2=kX/2=kX.0018

4.8=kX.0018

k=4.8/.0018=2666.7N/m