How much work is requried to uniformly accelerate a merry-go-round?
1 answer:
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Answer:
In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.
Explanation:
Hi there!
Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.
In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).
Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:
∑Fy = W + N = 0 ⇒ W = N
The weight is calculated as follows:
W = (mb + ms) · g
Where:
mb = mass of the bobsleder.
ms = mass of the sled.
g = acceleration due to gravity.
In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:
Fr = N · μ
Where:
N = normal force.
μ = kinetic friction coefficient.
Since N = W = (mb + ms) · g
Fr = (mb + ms) · g · μ
If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:
∑F = m · a
Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).
∑F = Fr = (mb + ms) · g · μ = (mb + ms) · a
(mb + ms) · g · μ = (mb + ms) · a
Solving for "a":
g · μ = a
