Answer: W = 294 J
Explanation: Solution:
Work is expressed as the product of force and the distance of the object.
W = Fd where F = mg
W= Fd
= mg d
= 15 kg ( 9.8 m/s²) ( 2m )
= 294 J
Answer:
uk = 0.25
Explanation:
Given:-
- An object comes to stop with acceleration, a = -2.45 m/s^2
Find:-
What is the coefficient of kinetic friction between the object and the floor?
Solution:-
- Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).
- We will apply equilibrium equation on the object in vertical direction.
N - m*g = 0
N = m*g
Where, N : Contact force exerted by the surface on the floor
g : Gravitational acceleration constant = 9.81 m/s^2
- Now apply Newton's second law of motion in the horizontal ( x-direction ):
- Ff = m*a
- The frictional force is related to contact force (N) by the following expression:
Ff = uk*N
- Substitute the 1st and 3rd expressions in the 2nd equation:
uk*m*g = -m*a
uk = a / g
- Plug in the values and solve for uk:
uk = - (-2.45) / 9.81
uk = 0.25
According to Hooke's Law formula. The force is proportional to the displacement of the spring. I believe
We got the following:
m = 78kg
h = 6m
g = 9.8 m/
≈ 10 m/
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A = ?
As we know A = Ep = mgh -> potential energy
so the answer would be A = 78 * 6 * 10 = 4680 Joule
