Question

Lemur and Chimp find a 200-kg treasure chest filled with gold plated bananas. The chest rests on an incline as illustrated below. The coefficient of friction between the chest and incline is 0.60. The incline makes a 35 degree angle with the horizontal. A cord parallel to the incline prevents the box from moving. A) What are the possible tensions in the cord? B) After Lemur cuts the cord, what is the magnitude of the acceleration of the treasure chest? C) What is the speed of the treasure chest after sliding 30 m down the incline?

Answer 1

Here if we see we have following problem statement:

force from gravity along inclined F_incline = m*g*sin(35) = 200 * 9.8 * 0.57 = 1124.21N

force from gravity perpendicular to incline F_perp = m*g*cos(35)

friction force is defined as coefficient of friction * perpedicular force by the ground to the mass(this perpendicular force is equal to the force applied by mass on ground).

so f_friction = coeff * m*g*cos(35)

coeff deifined in problem = 0.60

so f_friction = 0.60 * 200 * 9.8 * 0.82 = 963.32N

as the chord is also attached to the chest in parallel to the inclined and chest is at rest. That means the chord may have tension in it.

So tension T = Force from gravity along inclined - friction force

T = m*g*sin(35) - coeff*m*g*cos(35) = 1124.21 - 963.32 = 160.89N

if the chord is cut. then force which exist as the tension will act downward and make the block to move

so F_resultant = T = 160.89N

force = m* a

then accelaration of the block = F/m = 160.89/200 = 0.804 m/s²

now the chest initially was at rest so initial velocity = 0m/s

and total distance covered = 30 m

then final velocity is defined as follows:

v² = u² *2as, where u is initial velocity, a is accelaration and s is distance. putting all the values

v₂ = 0 + 2*0.804*30 = 48m²/s²

v = sqrt(48m²/s²) = 6.95m/s

Values required to solve the problem

A) tension in chord = 160.89N

B) accelaration = 0.804 m/s²

C) speed = 6.95 m/s