A team of slaves is dragging a heavy stone across the desert. The stone has a ma

Question

A team of slaves is dragging a heavy stone across the desert. The stone has a mass of 400 kg, and the force from the slaves is exerted horizontally, parallel to the ground. Between the stone and the ground, the coefficient of static friction is mu_s = 0. 5, coefficient of kinetic friction is mu_k = 0. 4. (a) The stone is sliding along with a velocity of 0. 1 m/s. If the slaves pull with a force of 1550 N, will the stone keep sliding with constant velocity? If not, what will be its acceleration? (b) The slaves stop for a coffee break, and when they start again, the stone is at rest. With what force must the slaves pull to get the stone to start moving again?

Explanation / Answer

Given:

Mass of the stone m = 400 kg

Coefficient of static friction µs = 0.5

Coefficient of kinetic friction µk = 0.4

Applied force in first scenario F = 1550 N

Initial velocity in first scenario v = 0.1 m/s

Acceleration of the block a = ? m/s2

Initial velocity in second scenario v’ = 0 m/s

Force required in second scenario F’ = ?

Solution

In the first scenario the stone is in motion during which the kinetic friction will resist the motion

Now the normal force (N) acting on the stone due to gravity

N = mg

N = 400 x 9.8

N = 3920 N

Now the kinetic friction Fk = µkN

Fk = 0.4 x 3920

Fk = 1568 N

Fk > F

The friction is resisting any kind of motion with a 1568 N force, so 1550 N force applied by the workers won’t be enough to cause a motion with a new acceleration, it will keep on sliding with a deceleration until the a certain distance (L) where all its kinetic energy got burned up as work

Net force Fn = F - Fk = - 18 N

a = Fn /m = -18/400 = - 0.045 m/s2

KE of the block

E = ½ mv2

= ½ x 400 x 0.12

=   2 J

Work done w = E

|Fn |L = 2 J

L = 2/18

= 0.11 m

In scenario b the stone is at rest during this time the static friction will resist the motion

The static friction Fs = µsN

= 0.5 x 3920

= 1960 N

So in order to create a movement the minimum required force is 1960N

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