Question

A 2 kg block is dragged up a hill at an acceleration of 1.8. If the hill is inclined at 30 and the coefficient of friction is 0.1 what force must be applied?

I got 15.2 but it was marked wrong

no other info

Answer 1

You must have done everything correct, only problem must have been with rounding off.

But just to make sure,

All the forces which you must consider are:

Gravitational forces which can be resolved into two (along the uncle and perpendicular to incline)

The force perpendicular to incline will be equal to the Normal force exerted on the block by incline.

There will be a frictional force directed down the incline, which will be equal to times the normal force.

And the force which is to be exerted will be given by addition of component of gravitational force (along the incline) & the frictional force and it's net force (m*a)

Component of gravity force along the incline = mgsin = 2*9.8*sin 30° = 2*9.8*(0.5) = 9.8N

Component of gravitational force (perpendicular to incline) = 2*9.8* cos30° = 2*9.8*(√3/2) = 16.97 N

Normal force = 16.97N (since normal force is equal to perpendicular component of gravitational force)

Now the magnitude of friction force= *N = 0.1*16.97 = 1.697 = 1.7 (approximately)

Net force on the block is = m*a = 2*1.8 = 3.6 N

So total force applied should be equal to ( mgsin + + m*a)

So now total force will be (9.8 + 1.7 + 3.6) = 15.1N