Question

A 2.55-kg object constrained to move along the x-axis is subjected to a time-varying force as shown on the graph Find the change in the a) Between 0 s and t4.23 s s velocity over the specified time intervals. Force vs. Time Number 15 A- 199 12 m/s b) Between 4.89 s and t 10s (N) Number .3 A26.4 m/ s .9 12 c) Between 4-10 s and t17.7s 0 24610 12 14 16 1820 t (s) Number Aw -22.4 1 m/s

A 2.55-kg object constrained to move along the x-axis is subjected to a time-varying force as shown on the graph below. Find the change in the object\'s velocity over the specified time intervals.

Answer 1

From the figure

The area under the force/time graph is the impulse:
impulse = change in momentum = Δp = F*Δt = m*Δv

a) Δp = 12N * 4.23s = 50.79 kg·m/s
so Δv = Δp / m = 50.79kg·m/s / 2.55kg = 19.9 m/s

b) You must make some assumption regarding the time where the force begins to decrease from 12 N. I'll assume t = 7.0 s. Then
Δp = 12N*(7.0 - 4.89)s + ½*12N*(10 - 7.0)s = 43.32kg·m/s
so p =m Δv
Δv=43.32/2.55
= 16.98 m/s

c) Δp = ½*-6N*(14-10)s - ½*(6+3)N*(17.7-14)s = -28.65 kg·m/s

so p =m Δv
Δv = -28.65/2.55
=-11.23 m/s