Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :

Horizontal component of force(Normal reaction) :

Since, box is on the verge of slipping :

Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
Answer:
7,546 J
Explanation:
recall that Potential energy is given by
P.E = mgΔh
where m = 70kg (given)
g = 9.8 m/s² (acceleration due to gravity)
Δh = change in height
= distance from top of building to top of car
= height of building - height of car
= (5+8) - 2
= 11m
substituting all these into the equation:
P.E = mgΔh
= 70 x 9.8 x 11
= 7,546 J
Answer:
5,970 N
Explanation:
m = 597 kg
a = 10 m/s^2
Plug those values into the following equation:
F = ma
F = (597 kg)(10 m/s^2)
F = 5,970 N