A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
friction between the block and the slope is 0.200. what is the acceleration of the block
1 answer:
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
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To solve this problem we will apply the concepts related to the balance of Forces, the centripetal Force and Newton's second law.
I will also attach a free body diagram that allows a better understanding of the problem.
For there to be a balance between weight and normal strength, these two must be equivalent to the centripetal Force, therefore
Here,
m = Net mass
= Angular velocity
r = Radius
W = Weight
N = Normal Force
The net mass is equivalent to
Then,
Replacing we have then,
Solving to find the angular velocity we have,
Therefore the angular velocity is 0.309rad/s