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8 15 10 5 10 Time in Secands The graph shows the linear momentum of a 1.8 kg object a) what is the object's displacement between t = 2 and t = 4 seconds? b) How much work is done upon the object between t 3 and t = 5.5 seconds? c) Calculate the net force acting on the object at t 1 second. Due no later than 8:20 a.m. on 11/08/2017 Physics 195 Fall 2017 Take-Home Quiz FOUR No work or explanation? No units?No boxed answer? No credit. 3) Calculation Question Answer on a new sheet of paper A 57 g tennis ball has velocity V-22(i) +1.2(-j) , and is 1 meter above the earth when it is struck by a tennis racquet moving in the opposite direction. As a result of the 0.25 s collision, the ball experiences an impulse of 3.2 Ns at an angle of 23 degrees from the horizontal. a) What is the range of the tennis ball? b) What is the average force experienced by the racquet during the collision? c) Calculate the net work done upon the tennis ball by the racquet during the collision. 4) Calculaion Question Answer on a new sheet of paper A system consists of 2 masses (6 kg and 4 kg) connected by a compressed spring (k 10,000 N/m spring compression distance 10 cm). The system is sliding across a level surface with a speed of 8 m/s. At t-0, the spring suddenly expands. a) Calculate the amount of work done upon each block in the center of mass frame b) Calculate the amount of work done upon each block in the earth frame c) Calculate the difference in the total kinetic energy of each frame. Why is there a difference? Support your response with calculations. Due no later than 8:20 a.m. on 1 1/08/2017

Explanation / Answer

4)

Given two masses

M = 6 kg

m = 4 kg

spring constant k = 10,000 N/m

spring compression, x = 0.1 m

speed u = 8 m/s

at t= 0, the spring expands

a.

Let the body M move by distance X and m move by distance x due to expansion of the psring

then x + X = 0.1 m

and as this is an internal force the position of center of mass relative to the masses will not change

MX = mx

6X = 4x

X = 4x/6

x + 4x/6 = 0.1

x = 0.06 m

X = 0.04 m

hence work done on the block m = 0.5kx^2 = 0.5*10,000*0.06^2 = 18 J

hence work done on the block M = 0.5kX^2 = 0.5*10,000*0.04^2 = 8 J

b.

In earth frame

initial energy of the system = 0.5(M + m)u^2 + 0.5k*0.1^2 = 0.5(10)8^2 + 0.5*10,000*0.1^2 = 370 J

when the psring is uncompressed, let speed of both the blocks be v

then from conservation of energy

0.5(M + m)v^2 = 370

v = 8.602 m/s

so work done on M = 0.5M(v^2 - u^2) = 30 J

work done on m = 0.5m(v^2 - u^2) = 20 J

c.

Difference in total of each frame = (30 + 20) - (18 + 8) = 24 J

this is because in center of mass frame the speed of center of mass also changes which is not considered in the center of mass frame

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