Posted by **Kailee** on Wednesday, July 2, 2014 at 10:42pm.

A trunk of mass 15 kg is on the floor. The

trunk has a very small initial speed.

The acceleration of gravity is 9.8 m/s2 .

What constant horizontal force pushing the

trunk is required to give it a velocity of 10 m/s

in 20 s if the coefficient of sliding friction

between the trunk and the floor is 0.58?

Answer in units of N

EXPLAIN ALSO??

ANOTHER QUESTION

1300 N a push of the box to the right

236 kg of the box

μ = 0.44

A crate is pushed horizontally with a force.

The acceleration of gravity is 9.8 m/s2 .

Calculate the acceleration of the crate.

Answer in units of m/s2

- physics -
**Henry**, Friday, July 4, 2014 at 5:48pm
1. Wt = m*g = 15kg * 9.8N/kg = 147 N. =

Weight of trunk.

Fk = u*mg = 0.58 * 147 = 85.26 N.=Force

of kinetic friction.

2. Fap = 1300 N.?

Mass = 236 kg

u = 0.44

Wc = m*g = 236kg * 9.8N/kg = 2313 N. =

Weight of crate.

Fk = u*mg = 0.44 * 2313 = 1018 N.

a=(Fap-Fk)/m=(1300-1018)/236=1.19 m/s^2

- physics -
**Henry**, Friday, July 4, 2014 at 5:52pm
1. The solution is incomplete.

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