A block is pushed up an inclined plane by a horizontal force f as shown in Figur
ID: 1421076 • Letter: A
Question
A block is pushed up an inclined plane by a horizontal force f as shown in Figure 7.P11. Attached to the upper end of the incline is a spring with stiffness constant k = 320. N/m. The horizontal force continues to act as the block moves an additional 0.500 m along the incline. Assume the following magnitudes for the forces acting on the block as it moves the 0.500 m after contacting the spring: P = 300. N, N = 260. N, W = 100. N, and F_fric = 70.0 N. Calculate the work done on the block by each of the following forces for the 0.500-m distance: (a) the push force P, (b) the normal force N, (c) the weight force W, (d) the frictional force F_fric, (e) the spring, (f) What is the net work done on the block over the 0.500-m distance? (g) If the block had a speed of 1.00 m/s when it first contacted the spring, what is its speed after traveling the 0.500 m as described above?Explanation / Answer
Work done = F.d = |F||d|cos@
where @ is the angle between force F and displacement d.
a) d = 0.5m along incline upward
P = 300 N horizontal
@ = 36.87 deg
W = 0.5 x 300 x cos36.87 =120 J
b) normal force is perpendicular to the inclien.
@ = 90 deg
W = 260 x 0.5 x cos90 = 0
c) W = 100 N
@ = 90 + 36.87 = 126.87 deg
Wrok done = 100 x 0.5 x cos126.87 = - 30 J
d) @ = 180 deg
Work done = 70 x 0.5 x cos180 = -35 J
e) work done by spring = - potential energy stored
= -kx^2 /2 = - 320 x 0.5^2 / 2
= - 40 J
f) Total work done = -40 -35 -30 + 0 + 120 = 15 J
g) using work energy theorem,
total work done = change in KE
15 = 10 ( vf^2 - 1^2) /2
vf^2 -1 = 3
vf = 2 m/s
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