How much work is done when a 2kg object is pulled upward?

No work is done. An approximately 2-kg object is pulled upward at constant speed by a 20-N force for a vertical displacement of 5 m. F grav and F tens do work. Forces do work when there is some component of force in the same or opposite direction of the displacement. 3.

How to calculate the amount of work done on an object?

The amount of work done upon an object depends upon the amount of force (F) causing the work, the displacement (d) experienced by the object during the work, and the angle (theta) between the force and the displacement vectors. The equation for work is W = F*d*cosine(theta)

What is the work done by gravity on a 15 kg box?

A 15 kg box falls at angle 25 ∘ from a height of 10 m. Determine the work done by gravity. Therefore, the work done by gravity is 1332 J.

What is the work done by gravity?

The work done by gravity formula is given by, W = mgh cos θ. W = 15 × 9.8 × 10 cos 25o. = 1470 J. Therefore, the work done by gravity is 1470 J. Example 2. A boy drags a 10 kg box across frictionless surface. He applies a force of 30 N at an angle of 32o to the horizontal for 6m.

What force is used to push a block across a friction surface?

A 10-N force is applied to push a block across a frictional surface at constant speed for a displacement of 5.0 m to the right. F app and F frict do work. F grav and F norm do not do work since a vertical force cannot cause a horizontal displacement.

Can a machine have a mechanical advantage greater than its velocity ratio?

It can have a mechanical advantage greater than the velocity ratio. Reason: If the mechanical advantage of a machine is greater than its velocity ratio, then it would mean that the efficiency of a machine is more than 100\%, which is practically not possible. A force of 5kgf is required to cut a metal sheet.

How do you calculate the amount of energy supplied to machine?

Energy supplied to machine (or input) = Effort × Displacement of the point of application of effort. Work obtained from machine (or output)= Load × Displacement of the point of application of load. For an ideal machine, Work output = Work input. (ii) To obtain gain in force. (iii) To change the direction of force.

How much force does it take to lift a 15-newton block?

To lift a 15-Newton block at constant speed, 15-N of force must be applied to it (Newton’s laws). Thus, 7. A student with a mass of 80.0 kg runs up three flights of stairs in 12.0 sec. The student has gone a vertical distance of 8.0 m. Determine the amount of work done by the student to elevate his body to this height.

What is the formula for the force of friction?

Fk = Force of kinetic friction, μk = coefficient of kinetic friction, N = Normal force or the force perpendicular to the contacting surfaces. μ s = Fs / N . Here, F = F s

How do you find the coefficient of friction at constant velocity?

At constant velocity to the Right, F = F k. μ k = Fk / N . Here, F = F k Fk = Force of kinetic friction, μk = coefficient of kinetic friction, N = Normal force or the force perpendicular to the contacting surfaces. μ s = Fs / N . Here, F = F s