A woman is weight training. She is doing back squats, where she places a barbell
ID: 1336548 • Letter: A
Question
A woman is weight training. She is doing back squats, where she places a barbell on her shoulders, squats down, and comes back up. Let’s only consider the upward part of the motion (starting from the squat). There are two parts to this motion, the initial acceleration, and the coasting towards the top. Suppose the leg muscles can exert a maximum force of 3000 N. To make it more simple, focus only on moving the barbell, and ignore the effort she has to put in to move herself.
a) If she exerts her maximum force to get started, how quickly would she accelerate a weight of 100 kg to her coasting speed of 20 cm/s?
b) How much force is required for the coasting part of the motion?
c) Suppose she decides to back off, and only exert about half as much effort. How quickly would she now accelerate to her coasting speed?
d) If she only exerts 1/3 as much effort, will she be able to stand up?
Explanation / Answer
a)
Force applies = 3000 N
Gravitational force acting downward = m*g = 100*9.8 =980 N
Net a = (net force)/mass = (3000 - 980)/100 = 20.2 m/s^2
vf = 20cm/s = 0.2 m/s
vi = 0
use:
Vf=vi + a*t
0.2 = 0 + 20.2 * t
t=0.0099 s
Answer: 0.0099 s
b)
In coasting part she must balance the weight
so, force required= m*g = 100*9.8 =980 N
Answer: 980 N
c)
Force applies = 3000/2 =1500 N
Gravitational force acting downward = m*g = 100*9.8 =980 N
Net a = (net force)/mass = (1500 - 980)/100 = 5.2 m/s^2
vf = 20cm/s = 0.2 m/s
vi = 0
use:
Vf=vi + a*t
0.2 = 0 + 5.2 * t
t=0.038 s
Answer: 0.038 s
d)
Force applies = 3000/3 =1000 N
Gravitational force acting downward = m*g = 100*9.8 =980 N
Net a = (net force)/mass = (1000 - 980)/100 = 0.2 m/s^2
Yes she can still get up but acceleration will be very small.
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