Question

2. A 10 kg block is dragged by an applied force of 20N, which makes an angle of 60⁰ with the horizontal. The coefficient of friction of the surface is. The block starts from rest and is pulled a distance of 7m.

1. Calculate the work done by the applied force?
2. Calculate the work done by the force of friction?
3. What is the resultant work done?
4. What is the final velocity of the block?

             Recall: work done equals change in kinetic energy                                  [4 pts.]

3. 4) On a frictionless surface, a block with mass M₁ =0.50 kg moving at 4 m/s to the right collides with a mass M₂= 0.20kg block moving to the left at 2m/s. The bodies move along the x-axis only.

a) If the two blocks stick together after the collision, what is the final velocity (magnitude and direction?

b) How much mechanical energy is lost?

c) If the collision lasts for 30ms, find the average force

Answer 1

F20N Swolian му кL Me = ? in your first question comefficient : of friction is not mentioned So, without (Mi) it is not possible to solve. amis - 4 amis mis E yzen a) | М, before. o.skg 0.2kg after from momentum conservatiom. m, U, + Mq U2 = (1, +Maj v 6-5x+ – 0-2x2 ( St 1 , con 2 -0.4 " ž 0-747 Tudla 12.28 mis Answer Scanned with CamScanner V I DA

b). initial mechanical energy Mi = k M, 4,? + f M2 V₂² = [X0.5 x 4² + 2 0.2 x 22 - 4 +0.4 = 4:49. Md = 1 {Mit Me) Vy? L CO.S+0.2) X(2-2812 =1.013 itt mis A Soy Loss - Mi-MH (on olosso = 4.4-1.81 Loss 1 2.50 57 Answer impulse = change in inomentom. Fit & Pf - Pir und hos la = 66.570257-[0-54/ 7.0.2x2] = 1.59.-: 1. 6A Fit 3-0.01 a) | F = 3.3N I Answer 0.01 30 X16-3 CScanned with CamScanner