Complete question is;
Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.
What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?
Answer:
F = 501.5 N
Explanation:
We are given;
Mass of wooden crate; m = 75 kg
Angle of ramp; θ = 11°
Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;
F = mgsin θ + μmg cos θ
From online values, coefficient of friction between wooden surfaces is μ = 0.5
Thus;
F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)
F = 501.5 N
Answer:
Explanation:
The net force on the potatoes is given by:
F= 52 - mgSintheta
F= 52- (2×9.8× Sin70°)
F = 52 -18.4
F= 33.58N
Using Newton's 2nd law
F = ma
a=F/m = 33.58/ 2 = 16.79m/s^2
Using the equation of motion:
V^2= u^2 + 2as
V^2 = 0 + 2× 16.79 x2
V^2 = 67.16
V=sqrt(68.16)
V= 8.195m/s This is the exit velocity of the potatoes
Kinetic energy, K.E = 1/2mv^2
KE= 1/2 × 2 × 8.195^2
KE = 67.16J
The planet is represented as Saturn ♄
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