A 2.00-kg mass slides down an incline as shown in Figure 8.P17. At the bottom of

Question

A 2.00-kg mass slides down an incline as shown in Figure 8.P17. At the bottom of the slope, it glides without friction along a horizontal surface and collides with a Hookes-law spring with stiffness constant k = 392 N/m. What is the potential energy of the mass at the top of the incline (relative to the bottom of the incline)? If the speed of the block at the bottom of the incline is 7.00 m/s, how much work did friction do on the block as it traveled down the incline? How far does the block compress the spring? (Assume the spring is massless.)

Explanation / Answer

part a )

U = mgh

U = 2 * 9.8 * 5 = 98 J

part b )

by energy conservation

if there is no friction then block speed at bottom around 10

U = KE + Wf

Wf = U - KE

Wf = mgh - 1/2 * 2 * 7^2

wf = 49 J

part c )

mgh = 1/2kx^2

x = sqrt(2mgh/k)

x = sqrt(2*2*9.8*h/392)

x = 0.707 m

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