Question

An inclined plane 20.0 m long has a slope of 40.0 degrees. An object with a mass
of 170. kg is sitting on the top of the incline. (a) If the coefficient of friction
between the mass and the plane is 0.235, find the velocity of the object at the
bottom of the incline. (b) If the block slides across a horizontal floor after
reaching the bottom of the incline, how far will the block travel if the coefficient
of friction remains 0.235.

I need serious guidance with this question. (Worksheet is due tomorrow.)

Answer 1

can u figure out the forces acting on the object parallel and perpendicular to the incline?

Answer 2

Would the forces be the force exerted and the weight of the object? @Rohitkhanna

Answer 3

the forces would be the weight of the object and the reaction by the incline

Answer 4

The reaction is...friction?

Answer 5

Or would it be g...9.8m/s^2? @Rohitkhanna

Answer 6

the reaction by the incline will have two components--
friction which will act parallel to incline and normal reaction which is perpendicular to the incline
the weight of the body will also have two components--one parallel to incline and other perpendicular to it

Answer 7

you need to find the net force acting on the object parallel to the incline to find its acceleration which will give its veloity at any moment.....
the net force on it perpendicular to the incline is zero as their is no movement perpendicular to incline

Answer 8

can u resolve the components of the weight(force) acting on the object perpendicular and parallel to it?

Answer 9

Yes, if I knew the correct formula to use. I keep forgetting.

Answer 10

ok.I will tell u

Answer 11

Wouldn't there be constant velocity involved in the problem? It says the friction remains the same from top to bottom.

Answer 12

the weight=mg acts vertically downward
its parallel component is given by mg x sin(angle)
and perpendicular component is mg x cos(angle)

Answer 13

constant velocity will be involved only if acceleration is zero

Answer 14

the normal reaction by the incline is equal to zero as net force is zero in that direction

firctional force will act up along the incline and it is given by--->friction coefficient x normal reaction

Answer 15

Hmm, would the equations be (170kg)(9.8m/s^2)sin(40degrees) and (170kg)(9.8m/s^2)cos(40degrees) .. ?

Answer 16

yes.the components found by u r correct

Answer 17

the net force parallel to incline acting on the object is
mg x sin(angle) - frictional force

Answer 18

after finding the net force,divide it by object's mass to get the acceleration

Answer 19

Parallel force = 1070.88 and perpendicular force = 1276.23 .. ?

Answer 20

@rohitkhanna 1070.88/170kg = parallel net force = acceleration...?

Answer 21

@rohitkhanna I got 6.3 for acceleration.