Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
Divide each side by (mass):
Acceleration = (force) / (mass)
= (100 N) / (50 kg)
= 2 m/s²
Answer:
W = 0.49 N
τ = 0.4851 Nm
Force
Explanation:
The weight force can be found as:
W = mg
W = (0.05 kg)(9.8 m/s²)
<u>W = 0.49 N</u>
The torque about the pivot can be found as:
τ = W*d
where,
τ = torque
d = distance between weight and pivot = 99 cm = 0.99 m
Therefore,
τ = (0.49 N)(0.99 m)
<u>τ = 0.4851 Nm</u>
The pivot exerts a <u>FORCE </u>on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.
