Question

2 0 2086420 10 86 2

A 250 g block is moving along the “x” axis and at some point a force in the positive direction of the axis acts on it.

The graph shows the magnitude of the force versus the position “x” of the particle. The curve is given by F=a/x^2, with a=9.0 N.m2 . Find the work done on the block by the force as the block moves from x=1.0 m to x=3.0 m by:

a.) Estimating the work from the graph.

b.) Integrating the force function.

When the force stops acting, at x=4.0 m, the block hits a relaxed horizontal spring, that has a spring constant of k=2.5N/cm. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping. The friction coefficient (µk) between the block and the surface is µk=0.25. (Yes, you have to consider friction). Consider that the friction starts acting after x=1 m.

a.) While the spring is being compressed, what work is done on the block by the spring force?

b.) What is the speed of the block just before it hits the spring?

c.) What is the speed of the block just before the force starts acting on it? The initial speed.

d.) If the speed of impact were doubled, what is the maximum compression of the spring?

Answer 1

a ) and b )

work done interms of force is

W = $\int_{1}^{3}$ F . dx

W = $\int_{1}^{3}$ a . x^-2 dx

W = $\int_{1}^{3}$ 9 X x^-2 dx

W = 9 X $\int_{1}^{3}$ x^-2 dx

W = 9 X ( x^-1 / (-1) ) limits 1 to 3

W = - 9 X ( 1/3 - 1)

W = - 9 X ( - 0.67 )

W = 6.03 J

next part solution

a)

work W = m g h

W = 0.25 X 9.8 X 0.12

W = 0.294 J

W = k x

W = 250 X 0.12

W = 30 J

b )

1/2 mv² + m g h = 1/2 k x²

0.5 X 0.25 X v² + 0.25 X 9.8 X 0.12 = 0.5 X 250X 0.12²

0.125 v² = 1.8 - 0.294

v² = 12.048

v = 3.47 m/s

c )

3.47² = u + 2 X 9.8 X 0.12

u = 12.048 - 2.352

u = 9.696 m/s

d )

1/2 m v² + m g h = 0.5 k x²

0.5 X 0.25 X 12.048 + 0.25 X 9.8 X x = 0.5 X 250 X x²

125 x² - 2.45 x - 1.50 = 0

x = 14.975 m