Determine the following (SHOW WORK STEP BY STEP) 1. Change in potential energy o
ID: 1282118 • Letter: D
Question
Determine the following (SHOW WORK STEP BY STEP)
1. Change in potential energy of the block as it slides down the ramp.
2. work done by gravity as the block slides down the ramp.
3. work done by friction as the block slides down the ramp.
4. net work done on the block as it slides down the ramp.
5. kinetic energy of the block at the bottom of the ramp.
6. work done by gravity as the block travels from B to C.
7. work done by the normal force as the block travels from B to C.
8. work done by friction as the block travels from B to C.
9. net work done on the block as it travels from B to C.
10. kinetic energy of the block at C.
11. work done by the spring as the block travels from C to D.
12.change in potential energy of the spring as the block travels from C to D.
A 5.0 kg block is released from rest (at point A) to slide down a 5.0 m ramp inclined at 30 degree (from A to B) then across a 10.0 m horizontal section (from B to C) until it collides with a spring (at C) wich it compresses by some amount before coming momentarily to rest (at D). Friction is known to act (k= 0.10) on the block as it travels from A to C. As the spring compresses (the block travels from C to D) there is no friction between the block and the horizontal ramp. Determine the following (SHOW WORK STEP BY STEP) 1. Change in potential energy of the block as it slides down the ramp. 2. work done by gravity as the block slides down the ramp. 3. work done by friction as the block slides down the ramp. 4. net work done on the block as it slides down the ramp. 5. kinetic energy of the block at the bottom of the ramp. 6. work done by gravity as the block travels from B to C. 7. work done by the normal force as the block travels from B to C. 8. work done by friction as the block travels from B to C. 9. net work done on the block as it travels from B to C. 10. kinetic energy of the block at C. 11. work done by the spring as the block travels from C to D. 12.change in potential energy of the spring as the block travels from C to D.Explanation / Answer
1. Change in PE is final PE-initial PE = 0 - mg*5sin(30) = -122.6J
2. WD by gravity is the thing which causes its potential energy to change. So same as 122.6J
3. WD by friction = Decrease in Total E of block
force balance: mgsin(30)-kmgcos(30)=ma. so acceleration the block experiences = a = g*sin(30)-kg*cos(30) where k is 0.1 (given). a= 4.05m/s2
Along AB: v^2=u^2+2as; u=0 as it starts from rest at A. Thus v^2 = 40.5 at B. Multiply by 0.5*m gives KE at bottom = 101.3J
Now KE at B is its Total E at B. Thus WD by friction = 101.3-122.6= -21.21 J (Negative because dissipative force)
4. Net work done on the block changes its Total Energy. Thus 21.21J
5. Pls refer 3
6. 0 as gravity and displacement are perpendicular to each other
7. 0 as Normal force and displacement are perpendicular to each other
8. Along BC frictional force is k*mg. Thus the deceleration the bloc experiences is kg= 0.981m/s2.
to get square of velocity at C (v): v^2=u^2-2*a*s where u^2=40.5 (refer 3). s =10m
thus v^2=20.88 at C.
WD by friction is the change in KE (PE =0 through out) = 0.5*m* (vC^2-vB^2) = -49.05 J
9. Net WD on the bloc is done by friction. (Same as above)
10. KE at C = 0.5*m*20.08= 50.2J
11. Spring force brings the block to rest. Thus its KE at C is reduced to 0 by spring. This is the WD by the spring as theres no friction. Ans = -50.2J since force and displacement are opposite.
12. Initially spriong was at rest. Now it is compressed. During this compression it reduces the speed of bloc to 0. This KE is stored in spring as its PE without any loss because of the absense of friction. Thus PE of spring = 50.2J
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