At t = 0 s, a wheeled cart, with a mass of 20.0 kg, has a velocity of 3.00 m/s i

Question

At t = 0 s, a wheeled cart, with a mass of 20.0 kg, has a velocity of 3.00 m/s in the negative x direction. The graph shows the net force applied to the cart, as a function of time, over the next 7.00 s. A positive force means the force is in the positive x- direction; a negative force means the force is in the negative x-direction.

a) What is the cart’s velocity at the end of the 7.00-second period?

b) What is the net work done on the cart during the 7.00-second period?

c) What is the maximum speed reached by the cart during the 7.00-second period? When does the cart reach this maximum speed?

d) Is the cart instantaneously at rest at any time(s) during this 7.00-second period? If so, when?

Explanation / Answer

a) dv/dt = F(t)/m

=> Final velocity after 7 seconds is, vf = vo + (1/m) 07 F(t)dt

Now,

for t = 0 to t = 1, F(t) = 40t

for t = 1 to t = 3, F(t) = 40

for t = 3 to t = 6, F(t) - 40 = [(-20 - 40)/(6 - 3)](t - 3)

which gives F(t) = -20t + 100

for t = 6 to t = 7, F(t) = -20

Using these values, we get,

vf = -3 + (1/20)[01 (40t)dt + 13 (40)dt + 36 (-20t + 100)dt + 67 (-20)dt]

=> vf = -3 + (1/20)[20 + 80 + 30 - 20]

=> vf = 2.5 m/s (in positive x-direction)

A faster way to calculate the integral is to find net area under the F-t graph.

b) According to Work-Energy Theorem:

KEi + Wnet = KEf

So, Wnet = KEf - KEi

=> Wnet = (m/2)(vf2 - vo2) = (20/2)[(2.5)2 - (-3)2] = -27.5 J

c) Maximum speed is either the initial speed or the speed at t = 5 s.

v5 = vo + (1/m) 05 F(t)dt

Now, 05 F(t)dt = Area under F-t graph from 0 to 5 seconds

=> 05 F(t)dt = [(1*40/2) + ((3 - 1)*40) + ((5 - 3)*40/2)] = 140 N/s

So, v5 = -3 + (1/20)*140 = 4 m/s

Since |v5| > |vo| , maximum speed = |v5| = 4 m/s

d) Since the velocity changed direction, it has to be zero somewhere. Let t be time at which the cart is instantaneously at rest.

Now, 0t F(t)dt = m(vt - vo) = 20(0 - (-3)) = 60 N/s

The area covered under the F-t graph from t = 0 to t = 1 is 20 N/s and the area under F-t graph from t = 1 to t = 3 is 80 N/s. So, velocity has to be zero somewhere between t = 1 and t = 3.

So, 20 + (t - 1)*40 = 60

=> t = 2 s

Hence, the cart is instantaneously at rest at t = 2 s

e) Net force is constant between t = 2 s and t = 3 s. So, cart's net displacement during this time is,

d = W23/Fnet = W23/40

where W23 = work done by the net force between t = 2 s and t = 3 s.

v2 = vo + (1/m) 02 F(t)dt = -3 + (1/20)[1*40/2 + (2-1)*40] = 0

v3 = v2 + (1/m) 23 F(t)dt = 0 + (1/20)[(3 - 2)*40] = 2 m/s

W23 = KE3 - KE2 = (m/2)[v32 - v22] = (20/2)[22 - 02] = 40 J

=> d = W23/40 = 40/40 = 1 m

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