work and power and forces

Question

1) Kelly is trying to deadlift a 68 kg barbell that is setting on the ground.

a) on her first attempt, she pulls straight upward with a force of 600N but the barbell does not move. How large is the normal contact force exerted by the ground on the barbell at the exact moment that kelly is pulling up on the barbell but it is not moving?

b) on her second attempt, kelly pulls upward on the same barbell with a force of 670N. What will the vertical acceleration of the kettlebell be as a result of this force?

Answer 1

(a) Since the barbell is at rest and not moving, the net force on it is zero. Therefore, the upward force applied by Kelly must be balanced by the downward force of the barbell on the ground, which we can call the normal force.

Thus, the normal force is 600 N, which is equal in magnitude and opposite in direction to the force applied by Kelly.

(b) The net force on the barbell is the difference between Kelly's upward force and the downward force of gravity on the barbell:

F_net = F_upward - F_gravity

where F_gravity is the weight of the barbell, given by:

F_gravity = m*g

where m is the mass of the barbell (68 kg) and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the given values, we get:

F_gravity = 68 kg * 9.8 m/s^2 = 666.4 N

Therefore,

F_net = 670 N - 666.4 N = 3.6 N

The vertical acceleration of the barbell can be found using Newton's second law:

F_net = m*a

where m is the mass of the barbell.

Substituting the given values, we get:

3.6 N = 68 kg * a

Solving for a, we get:

a = 0.053 m/s^2

Therefore, the vertical acceleration of the barbell due to Kelly's upward force is 0.053 m/s^2.