Wr = m*g = 0.5kg * 9.8N/kg = 4.9 N = Wt.
of the rat.
Fp = 4.9*sin34 = 2.740 N. = Force to
the plane.
Fn = 4.9*cos34 = 4.062 N. Normal = Force perpendicular to the plane.
Fk = u*Fn = 0.2 * 4.062 = 0.8125 N.
Fc-Fp-Fk = m*a
0.4-2.74-0.8125 = m*a
m*a = -3.15
a = -3.15/m = -3.15/0.5= -7.30 m/s^2.
The negative acceleration means that the rat is accelerating DOWN the plane
in spite of the upward force exerted by the cat. The opposing forces are greater than the force of the cat.
A rat with mass 0.5 kg is sliding down an inclined plane. The plane is at an angle of 34⁰ with respect to the ground, and the coefficient of friction between the rat and the plane is 0.2. Simultaneously, a cat is giving the rat a push UP the hill, with a force equal to 0.4 N. Is the rat accelerating down the hill? Show why or why not, using Newton’s second law. If the rat is accelerating, calculate his acceleration.
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