Physics Question?

Question

A 42 kg of ice slides down the plane incline 34 degrees in the figure. Assuming friction is negligible, what is the acceleration of the block down the incline? If the kinetic coefficient of friction is 0.060, what is the acceleration of the block down the incline?

Please show and explain all work. Please show the formula or formulas that were used.

Answer 1

The force acting on ice is
Fd=Wsin(34)
Fd=mg sin(34) =
Fd=230N
in case of friction
Total force F equal to Force down Fd less force of friction f
F=Fd-f where f= u W cos(34)

F= Wsin(34) - uWcos(34)
F=mg(sin(34) - u cos(34))
F=216N

Answer 2

Force on Ice block is fiven by F=Fx+Fy
Fx=mg sin(34)
Fy=mg cos(34) (since Fy is balanced by Normal force exterted by the plane on the block the only net force acting on the block going down an incline is Fx
Fx = 42 * 9.81 *sin(34)=230.4 N
F=ma >> a= F/m
a= 230.4/42 = 5.48 m/s.s

In case of friction the net force acting is given by F=Fx-Ff (-Ff because it is acting in the opposit direction to the motion)
Ff=u * Fn (mu_kinetic)*(Normal Force)
Fn=mg cos(34)=341.6N
Ff=u * Fn = 20.5 N
F = Fx - Ff = 230.4 - 20.5 =209.9 N
F= ma >> a = F/m = 209.9/42 = 4.99 m/s.s

Answer 3

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The plane incline = O =34 degrees

Mass of ice = m =42 kg

Weight of ice = w = mg =42*9.8 =411.6 N

When friction is negligible,component of weight parallel to plane = mgsinO

When friction is negligible,acceleration parallel to plane = g sinO=9.8*0.5592=5.48 m/s^2

When friction is negligible,acceleration parallel to plane = 5.48 m/s^2
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Normal reaction perpendicular to plane = R =mg cosO=411.6cos34=411.6*0.8290=341.23 N

Normal reaction perpendicular to plane =341.23 N

Friction = f = coefficient of friction(mu)*normal reaction(R)

Friction = f =(mu)*mgcos34

Resultant force down ward and parallel to the plane= P =mgsinO - f

Resultant force =P =mg[sin34 -(mu)cos34]

Acceleration due to friction = P /m

Acceleration due to friction = mg[sin34 -(mu)cos34 ] / m

Acceleration due to friction = g[sin34 -(mu)cos34]

Acceleration due to friction = 9.8 [0.5592 - 0.06*0.8290 ]

Acceleration due to friction = 9.8*0.5095

Acceleration due to friction = 4.9926 m/s^2

If the kinetic coefficient of friction is 0.060, the acceleration of the block down the incline is4.9926 m/s
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