A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on the top. A large loa
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Answer:
a.
Explanation:
The equation of the forces along the directions parallel and perpendicular to the slope are:
- Along the parallel direction:
where :
m = 6.0 kg is the mass of the box
g = 9.8 m/s^2 the acceleration of gravity
is the angle of the slope
is the coefficient of friction
R is the normal reaction
a is the acceleration
- Along the perpendicular direction:
From the 2nd equation, we get an expression for the reaction force:
And substituting into the 1st equation, we can find the acceleration:
Solving for a,
Answer:
part (a)
Part (b)
Explanation:
Given,
- mass of the smaller disk =
- Radius of the smaller disk =
- mass of the larger disk =
- Radius of the larger disk =
- mass of the hanging block = m = 1.60 kg
Let I be the moment of inertia of the both disk after the welding,
part (a)
A block of mass m is hanging on the smaller disk,
From the f.b.d. of the block,
Let 'a' be the acceleration of the block and 'T' be the tension in the string.
Net torque on the smaller disk,
From eqn (1) and (2), we get,
part (b)
In this case the mass is rapped on the larger disk,
From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,
Let '' be the acceleration of the block in the second case,
From the above expression,
