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## Homework Statement

A wedge of mass m= 1kg that is initially at rest is driven up a friction less ramp by a horizontal force

F=50Ndue to a small attached rocket. The ramp is equilateral, with sides of length 50m

What is the magnitude of the normal force exerted by the ramp on the wedge?

What is the speed Vt (top of the ramp i assume) of the wedge as itl eaves the ramp?

## Homework Equations

F=ma

Fx=mgsin(60)

Fnety=n-mgcos(60)

(Final velocity)^2 – (Initial velocity)^2 = 2 * acceleration * distance

## The Attempt at a Solution

So i know an equilateral triangle has angles of 60*

The normal force of the block on the incline is:

F(normal) = mg*cosθ

= (1)(9.8)cos(60°)

= 4.9.

(b) The force of gravity acting on the block on the incline is:

F(gravity) = mg*sinθ

= (1)(9.8)sin(60°)

= 8.487

This implies that the net force is:

F(net) = F(push) - P(graivty)

= 4.9-8.487

= -3.58 N. Is this the same as the magnitude?

Using F = ma ==> a = F/m, the acceleration of the block is:

a = F/m

= -3.58/1

= -3.58 m/s^2 (down the incline).

Is this more along the right track?

If i have the acceleration of -3.58 i can just find the final velocity with (Vf)^2 – (Vi)^2 = 2(a)(d)=18.9m/s

I just want to make sure i have the right idea