The figure is attached below.

a) The forces acting on the cart are the thrust force along the incline from left to right `F_t` , the normal of the plane on the cart (the force that is sustaining the cart on the plane) `N` , the weight of the cart...

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The figure is attached below.

a) The forces acting on the cart are the thrust force along the incline from left to right `F_t` , the normal of the plane on the cart (the force that is sustaining the cart on the plane) `N` , the weight of the cart `G` directed downwards and the inertia of the cart itself `m*a` . Usually on a free body diagrsm the inertia force is not shown.

b) Since the forces `F_t` , `N` and `G` that act on the cart are constant in magnitude then the acceleration of the cart needs to be constant. Written as vectors the following equation holds:

` `m*a =Ft + N + G

c) The net force on the cart is

`F = m*a`

Assuming the mass of the cart is, for example `m=10 kg` then the total force on the cart is

`F =10*(-0.74) =-7.4 N`

The minus sign comes from the fact that the cart is moving upwards accelerated on the ramp.

d)

The weight of the cart can be decomposed along the inclined and perpendicular to the inclined.

`G_n = G*cos(alpha)` and `G_p=G*sin(alpha)`

The angle of the inclined comes from

`m*a +G_p =F_t`

`m*a +m*g*sin(alpha) =F_t`

`alpha = arcsin[(F_t-m*a)/(m*g)]`