A cart of mass 380 grams is temporarily held stationary on an incline?

Question

A cart of mass 380 grams is temporarily held stationary on an incline at 3.7 degrees to horizontal.
•A mass of 75.00001 grams is attached to the cart by a string over a pulley located at the bottom of the incline; the mass hangs freely from the string.
•When the cart is released, what will be its acceleration in the direction down the incline?
•Sketch and label a reasonable facsimile of a scaled diagram showing the forces acting on the cart.

Answer 1

The force responsible for the cart to move down is F ins due to the projection of its weight w1 onto the inclined plane.

F=w1 sin(3.7)
w1=m1 g
F=m1 g sin(3.7)

The force preventing the cart to move down the incline is the weight w2 of the mass m2 attached over the pulley.

w2=m2 g
Total force is thus
Ft=F-w2
Ft=m1 g sin(3.7) - m2g
Ft=(m1sin(3.7) - m2)g

also Ft=(m1 + m2)a

then a=Ft/(m1+m2)

a = (m1sin(3.7) - m2)g/(m1+m2) putting it all in kg--m-s units we can compute acceleration a
a=(.380 sin(3.7) -0.075) 9.81/(.380 + .075)
a=-1.09 m/sec^2 (note the negative sign. The cart will move up not down the incline)

Answer 2

The coin could have greater acceleration and could attain the backside first. because of the fact coin has smaller 2d of inertia, much less power is going into its rotation, and extra into translational action while in comparison with the hoop.