Ablock of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. a force f of magnitude 5.0 n is applied to the block in the horizontal direction (not parallel to the ramp).a) if this force is just large enough to start the block moving down the ramp, what is the coefficient of static friction μsbetween the block and the ramp? b) if the coefficient of kinetic friction between the block and the ramp has a value of 0.40, what is the magnitude of the acceleration of the block along the ramp as this force is applied to the block? c) what is the work done by this applied force if the block travels a distance of 2.0 m down (i. e., parallel to) the ramp?

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y