Help with Part #2 (solve and show work), please! A.10 kg box, initially at rest,

Question

Help with Part #2 (solve and show work), please!

A.10 kg box, initially at rest, moves along a friction less horizontal surface. A horizontal force to the right is applied to the box. The magnitude of the force changes as a function of time as shown. Rank the impulse applied to the by this force to the box by this force during each 2-second interval indicated below. Explain your reasoning. Assume no other horizontal forces act on the box. If the box was initially moving at 5.0 m/s to the left at t = 0.s. what is its speed at (show your work!) r = 2.0s r = 4.0 s r = 6.0 s r = 8.0 s t = 10.0 s

Explanation / Answer

you have found the impulse in the first part

so

we know Impulse = m delta v

impulse = Fdt

Fdt = m dv

foe the interval

0-2s

F= t

so the equation becomes

tdt = m dv

= integrating we get

= t^2 / 2 = m[v -v0]

= at t= 2 sec

= v = 5+ [0.2]= 5.2 m/s

2-4 s

F= 2t

2tdt = m dv

= t^2 = m[v-v0]

= at t= 4 sec

= 1.6 + 5 = v = 5.6 m/s

at 5-8 s

F = -t

so -tdt = mdv

= -t^2/2 = m[v-vo]

= at t= 6 s

= 5-1.8 = v

= 3.2 m/s

at t = 8s

-t^2 /2 = m [v-v0]

= 1.8 m/s = v

so now the force is 0 for 8-10 s

so it moves with a constant velocity

so at 10s v = 1.8 m/s

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