Question

A 7.75 kg(75.95 N)box is placed on a 29.7° incline on top of a lab table. The ramp is 3.11 m long, at the bottom of the incline the box slides horizontally for a short distance before rolling off the lab table. If the table is 0.84 m high, find how far from the table the ball will land. With the friction being 5.88 i found acceleration to be 4.1m/s2. The final velocity at the bottom of the plane if it started from rest is 25.50m/s.

Answer 1

m*g = 75.95 N. = Wt. of the box.

Fp = 75.95*sin29.7 = 37.63 N. = Force
parallel to the ramp.

a = (Fp-Fk)/m
a = (37.63-5.88)/7.75 = 4.1 m/s^2

V^2 = Vo^2 + 2a*d = 0 + 8.2*3.11 = 25.5
V= 5.05 m/s = Xo = Initial hor. velocity.

h = 0.5g*t^2 = 0.84 m.
4.9*t^2 = 0.84
t^2 = 0.171
Tf = 0.414 s. = Fall time.

Dx = Xo*Tf = 5.05m/s * 0.414s = 2.1 m.
From table.