Answer:
W = 7.06 J
Explanation:
From the given information the spring constant 'k' can be calculated using the Hooke's Law.
Now, using this spring constant the additional work required by F to stretch the spring can be found.
The work energy theorem tells us that the work done on the spring is equal to the change in the energy. Therefore,
![W = U_2 - U_1\\W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}(275.13)[0.29^2 - 0.18^2] = 7.06~J](https://tex.z-dn.net/?f=W%20%3D%20U_2%20-%20U_1%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx_2%5E2%20-%20%5Cfrac%7B1%7D%7B2%7Dkx_1%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28275.13%29%5B0.29%5E2%20-%200.18%5E2%5D%20%3D%207.06~J)
He has a mass of 56 kg.
The equation given is PE = mgh.
PE = 4620 J
h = 8.4
g = 9.8
Therefore:
4620 = 82.32m
m = 4620/82.32
m = 56 (rounded to two significant digits)
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
