Question

A block of mass 3.5 kg, sliding on a horizontal plane, is released with a velocity of 1.6 m/s. The block slides and stops at a distance of 1.6 m beyond the point where it was released. How far would the block have slid if its initial velocity were increased by a factor of 3.5

Answer 1

Answer:

19.6 m

Explanation:

The work-energy theorem applies here,

The theorem states that the change in momentum of a particle between two points is equal to the work done in moving the force between the two distance.

ΔK.E = W

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Kinetic energy = (1/2)(m)(v²)

m = 3.5 kg, v = 1.6 m/s

Final kinetic energy = 0 J, since the block of mass comes to rest.

Initial kinetic energy = (1/2)(3.5)(1.6²) = 4.48 J

ΔK.E = - 4.48 J

The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement.

W = - Fd = - 1.6 F

ΔK.E = W

- 4.48 = - 1.6 F

F = 2.8 N

when the initial velocity is increased by a factor 3.5,

v = 1.6×3.5 = 5.6 m/s

Final kinetic energy = 0 J, since the block of mass comes to rest.

Initial kinetic energy = (1/2)(3.5)(5.6²) = 4.48 J

ΔK.E = - 54.88 J

The workdone on the block of mass is done by the frictional force, F, which acts opposite to the direction of the displacement. The frictional force is the same as above.

W = - Fd = - (2.8) d

ΔK.E = W

- 54.88 = - 2.8 d

d = 19.6 m

Compared to 1.6 m, 19.6 m is an increase by a factor 12.25.