Solution - American Association of Physics Teachers
Solution - American Association of Physics Teachers Solution - American Association of Physics Teachers
2013 F = ma Exam 2025. A box with weight W will slide down a 30 ◦ incline at constant speed under the influence of gravity and frictionalone. If instead a horizontal force P is applied to the box, the box can be made to move up the ramp at constantspeed. What is the magnitude of P ?(A) P = W/2(B) P = 2W/ √ 3(C) P = W(D) P = √ 3W(E) P = 2WSolutionIf block slides down at constant speed, thentan θ = µIf block is pushed by horizontal force P up ramp at constant speed, thenF N = mg cos θ + P sin θso friction isF f = µ(mg cos θ + P sin θ)This is balanced by gravity and the push up the ramp, soP cos θ = µ(mg cos θ + P sin θ) + mg sin θDivide through by cos θ, and use first equationP = µ(W + P µ) + W µororP = 2µ1 − µ 2 WP = 2 √ 1 31 − 1 W3Copyright c○2013 American Association of Physics Teachers
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2013 F = ma Exam 2025. A box with weight W will slide down a 30 ◦ incline at constant speed under the influence <strong>of</strong> gravity and frictionalone. If instead a horizontal force P is applied to the box, the box can be made to move up the ramp at constantspeed. What is the magnitude <strong>of</strong> P ?(A) P = W/2(B) P = 2W/ √ 3(C) P = W(D) P = √ 3W(E) P = 2W<strong>Solution</strong>If block slides down at constant speed, thentan θ = µIf block is pushed by horizontal force P up ramp at constant speed, thenF N = mg cos θ + P sin θso friction isF f = µ(mg cos θ + P sin θ)This is balanced by gravity and the push up the ramp, soP cos θ = µ(mg cos θ + P sin θ) + mg sin θDivide through by cos θ, and use first equationP = µ(W + P µ) + W µororP = 2µ1 − µ 2 WP = 2 √ 1 31 − 1 W3Copyright c○2013 <strong>American</strong> <strong>Association</strong> <strong>of</strong> <strong>Physics</strong> <strong>Teachers</strong>