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
You might be interested in
Answer:
b) √[(kx²/m) - 2gx]
Explanation:
The energy at the lowest point is equal to:
where:
Eelas = elastic energy [J]
k = spring constant [N/m]
x = extension of the spring [m]
We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.