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
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
2 answers
1. The solution is incomplete.
2 answers