A block of mass 2kg is moving along the x-axis under the influence of a force wh

Question

A block of mass 2kg is moving along the x-axis under the influence of a force whose x-component varies as shown in the graph. The particle is at rest when it is at the origin. Determine the work done by the force as the block moves from x=0 to x= 4m from x=0 to x=8m Find the speed of the block at x = 4m. the speed of the block at x = 8m. A particle moving along the positive x-axis is attracted to the origin by a force that varies as, where C is a positive constant. Calculate the work done by this force on the particle as it moves from y_1 to y_2. What is the sign of the work if y_1 > y_2. What is the sign of the work if y_1 > y_2?

Explanation / Answer

Here ,

mass , m = 2 Kg

a) work done by the force is given as

work done = area under the curve

work done from x = 0 to x = 4 m

work done from x = 0 to x = 4 m = 0.5 * 30 * 2 + 30 * 2

work done from x = 0 to x = 4 m = 90 J

the work done from x = 0 to x = 4 m is 90 J

b) from x = 0 to 8 m

total work done = 0.5 * 30 * 2 + 30 * 2 - 20 * 2

total work done = 90 - 40

total work done = 50 J

the total work done from x = 0 to x = 8 m is 50 J

c)

for the speed of block is v m/s

Using work energy theorum

0.5 * 2 * (v^2 - 0) = 90

solving for v

v = 9.47 m/s

the speed block is 9.47 m/s

d)

for the speed at x = 8 m

Using work energy theorum

0.5 * 2 * (v^2 - 0) = 50

solving for v

v = 7.07 m/s

the speed block is 7.07 m/s

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