Work done by particle under application of an external force=∫Fdx=Area under the F vs x curve= 21​ ×(4+12)×10J=80JThis work done goes into the kinetic energy of the particleThus 21​ mv 2 =80⟹ 21​ ×0.1×v 2 =80⟹v=40m/s

The work done on a particle enacted to move from (0,0)(0,0) to (2,4)(2,4) along the parabola y=x2𝑦=𝑥2 by the force F––(t)=x2i–+y2j–𝐹_(𝑡)=𝑥2𝑖_+𝑦2𝑗_ is given by

A mass of 8 kg is traveling at a constant 10.4 m/s when a 38 N force, F, starts to act in the opposite direction, stopping the object after it has traveled. How much work was done by the force, F, in stopping the object?

In the figure provided, a particle is travelling along a closed vertical path. If the particle is starting and finishing at point A, how much work is done on the particle by gravity? 9.8 J 4.9 J 0 J 12.0 J

The Gravitational Field In a region is given by E = (3i-4j) N/kg Find the work done (in joule) in Displacing a particle by 1m along the line 4y= 3x+9.

Certain force acting on a 20 kg mass changes its velocity from5 m s –1 to 2 m s –1 . Calculate the work done by the force