Question

1-4

using Work-Energy Theorem/KE and PE

thank you!

not 1. An object of mass 0.5kg is projected from ground at 40m/s. Find the work done by gravity on the object, when it is at (a) 10m above the ground -49 Joules (b) 15m above the ground.- 73.5 Joules (c) Find the potential energy of the object in (a) and (b) relative to the ground. PE-mgh 49 J/73.55 2. A block of is sliding down a slope as shown in the diagram. When the block is 10m above the reference it's speed is 12m/s. m= 2 kg hi=10m Vi= 12mls hea 5m 10m th= 5m 2kg d=7m Sih Zm KE: d= 7m PE: (a) Find the initial kinetic energy and the potential energy of the block 70.6 J 1 9 J I (b) what is the speed of the block when it is at 5m above the reference? Ignore friction. 8.4 mis (c) If there is a constant friction force of 4N between the block and the slope what is the speed of the block at 5m? 2(12) 3. An object of mass 2kg moving at 6m/s, collides with a stationary object of mass 3kg. Find the speeds of the objects after the collision if (a) The collision is elastic (b) completely inelastic P=mov 4. A tennis ball of mass 0.2kg strikes a floor at 12m/s and bounces off at the same speed. The speed of the ball makes a 90° angle with the floor. (a) Find the momenta of the ball before and after the collision? (a) Find the Impulse exerted on the floor (b) If the ball is in contact with the floor for 0.1s, what is the average force exerted on the floor?

Answer 1

Given m= 0.5kg; initial speed u=40mis. - Here, determine the work done by the gravity on the object. - According to work energy theorem; work done is given by the change in kinetic enengy and also work done is canal to negative change in potential energy. W = AKE = - APE on w= Kg-ki = P.-P₂ when the object at lom above the ground (harom. Change in potential energy, 0 Initial potential energy P = mg(o)=0 cet h=10m, h = ng xh = 0.5*9.8 x 10 J So, work done w - Pink W =-0.5x9.8 x 10 J = - 49 J. - Em ).

© At h=15m. Po= mg.o =0 Pa = mghao.jny x 9.8 m/s2 x 15m 73.5 J work done wap , -P2 = 0-73.5J - woe - 73.5J Ans). is In @ The potential energy relative to the ground . P = mgh = 0.5ug x 9.8 m L x com 1. Pa = +49J (Hene p is the) In ① The P. E relative to ground is - Po= mgh = 0.5x9.8 x 155 Bu = + 73.55 (Tue) (an) I As Pen Chegy guidelines, i can do only question or mention a perticular I cuestion. Please post one which you want. individual I