1. on the slide, initial PE (mgh) equals final KE (1/2 mv^2)
2. work done on the friction=KE lost, or 1/2 mv^2 where v is the final velocity at the bottom of the slide
3. How far?
work done by friction= Ff*distance= mg*mu*distance
and that equals work done on friction
mg*mu*distance= KE at bottom of slide solve for distance
How can I solve this problem?
A block of mass 75 kg is released from rest at the top of a smooth frictionless incline from an initial height of 2 m. At the bottom of the incline, the block slides along a rough horizontal surface until it stops. The coefficient of kinetic friction between the rough surface and the block is u = 0.35.
1. How fast is the block moving when it reaches the bottom of the incline?
2. How much work was done by the friction force during the sliding motion on the rough horizontal surface?
3. How far does the block slide on the rough surface before coming to a stop?
How do I solve this and what formulas should be used?
2 answers
1. V^2 = Vo^2 + 2g*h = 0 + 19.6*2 = 39.2.
V = 6.26 m/s.
2. Work = change in KE = 0.5M*V^2 = 0.5*75*6.26^2 = 1470 Joules.
3. Fk = u * Fn = u * M*g = 0.35 * 75 * 9.8 = 257.3 J. = Force of kinetic friction.
Work = Fk * d = 1470.
257.3 * d = 1470,
V = 6.26 m/s.
2. Work = change in KE = 0.5M*V^2 = 0.5*75*6.26^2 = 1470 Joules.
3. Fk = u * Fn = u * M*g = 0.35 * 75 * 9.8 = 257.3 J. = Force of kinetic friction.
Work = Fk * d = 1470.
257.3 * d = 1470,
1 answer