Work, Friction, Inclined Plane

[mahi]

If a mass of 100kg is to be pushed up a plane inclined at 20 degrees from the horizontal, with a total displacement of 2.0 m, and a coefficient of friction of 0.20, how much work has to be done? Looking for the formula. Thanks!

 

[Physic_Scholar]

Work is defined as force (parallel) component multiplied by the distance.Remember to label all the forces acting on this object of mass 100g.

 

[quicknote]

I would like to first clarify that work is defined as the product of force and displacement, and is measured in joules (J). In this scenario, the force being applied is the force required to overcome both the weight of the mass (mg) and the force of friction (Ff). The formula for calculating work is W = Fd, where W is work, F is force, and d is displacement.

To find the total force required to push the mass up the inclined plane, we can use the formula F = mg sinθ, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of inclination (20 degrees in this case). Plugging in the values, we get F = (100 kg)(9.8 m/s^2)(sin 20) = 333.2 N.

Next, we need to calculate the force of friction (Ff) using the formula Ff = μN, where μ is the coefficient of friction and N is the normal force (perpendicular to the surface of the inclined plane). The normal force can be calculated as N = mg cosθ, where θ is the angle of inclination (20 degrees in this case). Plugging in the values, we get N = (100 kg)(9.8 m/s^2)(cos 20) = 940.6 N. Therefore, Ff = (0.20)(940.6 N) = 188.1 N.

To find the total work required, we can now use the formula W = Fd, where d is the total displacement (2.0 m in this case). Plugging in the values, we get W = (333.2 N + 188.1 N)(2.0 m) = 1042.6 J.

In summary, the total work required to push a mass of 100kg up a plane inclined at 20 degrees from the horizontal, with a total displacement of 2.0 m and a coefficient of friction of 0.20, is 1042.6 J.