Answer
given,
force = 94 lb
weight of crate = 220 lb
Assuming the static friction be equal = 0.47
kinetic friction = 0.36
Maximum force applied to move the object is when object is just start to move.
F = μ N
F = 0.47 x 220
F = 103.4 lb
As the frictional force is more than applied then the object will not move.
so, the friction force will be equal to the force applied on the object that is equal to 94 lb.
hence, the direction of force will left.
For each load, Work = (mass) x (gravity) x (distance .
Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .
Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.
The work required is equal in both cases.
The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .
If you mean the SI Unit of GPE, the answer is J for Joules.
if that's not the question being asked, i would need a little more elaboration please :)
Answer:
ac = 3.92 m/s²
Explanation:
In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,
Frictional Force = Centripetal Force
where,
Frictional Force = μ(Normal Force) = μ(weight) = μmg
Centripetal Force = (m)(ac)
Therefore,
μmg = (m)(ac)
ac = μg
where,
ac = magnitude of centripetal acceleration of car = ?
μ = coefficient of friction of tires (kinetic) = 0.4
g = 9.8 m/s²
Therefore,
ac = (0.4)(9.8 m/s²)
<u>ac = 3.92 m/s²</u>
