An initially stationary 1.5 kg object accelerates horizontally and uniformly to
ID: 1433575 • Letter: A
Question
An initially stationary 1.5 kg object accelerates horizontally and uniformly to a speed of 12 m/s in 2.4 s. (a) In that 2.4 s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval? At a certain instant, a particle-like object is acted on by a force F = (5.3 N )i - (3.1 N )j + (5.2 N )k while the object's velocity is v = - (3.3 m/s )i + (5.3 m/s )k . what is the instantaneous rate at which the force does work on the object? At some other time, the velocity consists of only a y component. If the force is unchanged, and the instantaneous power is -5.7 W, what is the velocity of the object just then? (Give your answer without a unit vector.)Explanation / Answer
F = ma
W = F * d
P = dW/dt = F * v
a = dv/dt = 12 m/s / 2.4 s = 5 m/s^2
For the work done there are two ways
1)
x(t) = 0.5 a t^2 + v0 t + x0
a = 5 m/s^2, v0 = 0, x0 = 0
x(2.4) = 14.4 m
W = 1.5 kg * 5 m/s^2 * 14.4 = 108 J
2)
by kinetic energy
k.e. = 0.5 m v^2 = 108 J
The force is constant so the power is the F * v.
1.5 * 5 * 12 = 90 W
1.4 * 5 * 6 = 45 W
x(2.4)
2)
a)
the power or work per unit time is
p = dw/dt
p = d/dt(F.v)
p = Fdd/dt
p = F.v
p = (5.3i - 3.1j + 5.2k).(-3.3i + 5.3k)
p = (5.3 x (-3.3)) + 5.2 x 5.3
p = 10.07 watts or J/s
b)
if F stay the same V = Voj with p = - 5.7 w
p = FyVo
-5.7 = -3.1 x Vo
Vo = 1.84 m/s