An engine provides a constant force F to a car of mass M that carries a box of m
ID: 2076617 • Letter: A
Question
An engine provides a constant force F to a car of mass M that carries a box of mass m on the top of it. After traveling from rest with constant acceleration a1 for 20 seconds, the speed of the car becomes 40 m/s and at the instant, the box falls off the car. In the next 20 seconds, with force F remaining constant, the car travels with constant acceleration, a2, 1280 meters from the point the box fell.
a. The ratio of the acceleration a2/a1 would be
It should be 3.2, 2.0, 1.2, or 1.0. Please show work.
b. The ratio m/M would be
It should be a2/a1, (a1+a2)/a2, (a2-a1)/a1, or (a2+a1)/a1. Please show work. Thanks
Explanation / Answer
FOR first 20 sec,
v = u + a t
40 = 0 + 20a1
a1 = 2 m/s^2
after thta for 20 sec,
d = v0 t + a t^2 /2
1280 = (40 x 20) + (a2 x 20^2 / 2)
a2 = 2.4 m/s^2
(A) a2/ a1 = 2.4 / 2 = 1.2 ........Ans
(B) F = (M + m) a1
M + m = F / a1 ..... (i)
and F = M a2
M = F / a2 ......(ii)
(i) - (ii) => m = F(1/a1 - 1/a2) = F (a2 - a1) / (a1 a2 ) .....(iii)
m / M = (iii) / ii = [ F (a2 - a1) / a1 a2 ] / (F / a2)
= a2 - a1 / a1 .........Ans