Due to application of 5 N force an object moves 10 meter along perpendicular direction of the force. What amount work is done?
Correct answer: D. 0 J
- A. 50 J
- B. 5 J
- C. 15 J
- D. 0 J
Explanation
The explanation of this question is: Work is a scalar quantity that can be represented as a dot product of 2 vector quantities force and displacement. Or W = fd cos θ where θ = Angle between force and displacement When force and displacement are perpendicular, then θ = 90o and cos 90o = 0 so W = fd x 0 = 0
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Work
Work is done when a force produces displacement, with its value determined by the component of force along the displacement. The topic covers positive, negative and zero work, work by variable forces from force-displacement graphs, and the connection between net work and change in kinetic energy.
Practise Work and Energy
750 free Work and Energy MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Work questions
A 50kg man with a 20kg load on his head climbs up 20 steps of 0.25m height each. The work done in climbing is?
The adjacent figure shows the force-displacement graph of a moving body, the work done in displacing the body from x=0 to x=35m is equal to:
A force of 20 N acts on a body through a distance of 10m. What must be the angle between force and displacement such that work comes out to be 100 J?
A crate moves 10m to the right on a horizontal surface as a woman pulls on it with a 10N force. Rank the situations shown below according to the work done by her force from least to greatest.
A man moves the 10 g object shown in a vertical plane from position "x" to position "y" along a circular track of radius 20 m. The process takes 0.75 min. The work done by the man is about: (g = 10m/s2)
If a body of mass 2kg is raised vertically through 2m, then the work done will be: