A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to
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Answer:
T=13.72N
Explanation:
The tension before the ball is released have no angle is in rest at the same axis of the weight so:
∑F=0
Using Newton law in this case the ball is tied so tension before become to swing is
∑F=FN-T=0
1. 12.75 J
Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is
where
F = 15 N is the magnitude of the force
d = 85 cm = 0.85 m is the displacement of the cart
Substituting in the formula, we get
2. 10.6 N
In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).
Therefore, the work done is
However, in this case the displacement is
d = 120 cm = 1.20 m
Therefore, the magnitude of the force in this case is
When something moves on a round track, the guidance of the something's velocity must continually switch. A switching velocity means that there must be an acceleration. This acceleration is horizontal to the guidance of the velocity. This is said as “the radial acceleration”, or “centripetal acceleration” ("centripetal" means "center searching"). The “radial acceleration” is equal to “the square of the velocity”, divided by “the radius of the circular path of the object”. The unit of the “centripetal acceleration” is m/s².
where,
"v" = "velocity" (m/s) and "r" = "radius of motion of the object" (m)