A block of mass 10kg is pulled horizontally by a force of 75N that makes an angle of 37° to the horizontal whose coefficient of friction is 0.4 . what is the acceleration of the block
M*g = 10kg * 9.8N/kg = 98 N. = Wt. of
block.
Fn = M*g - Fap*sinA = 98 - 75*sin37 =
52.9 N.
Fk = u*Fn = 0.4 * 52.9 = 21.15 N. = Force of kinetic friction.
a = (Fap*Cos37-Fk)/M =
(75*Cos37-21.15)/10 =
To find the acceleration of the block, we need to analyze the forces acting on it. There are three main forces at play: the applied force, the force of friction, and the force of gravity.
1. Applied Force: The force pulling the block horizontally is given as 75N at an angle of 37° to the horizontal. We need to find the horizontal component of this force. Using the formula:
F_horizontal = F_applied * cos(angle)
F_horizontal = 75N * cos(37°)
F_horizontal = 75N * 0.7986
F_horizontal = 59.895N (rounded to three decimal places)
2. Force of Friction: The force of friction opposes the motion of the block. It is given by:
F_friction = coefficient of friction * normal force
The normal force is equal to the weight of the block, which is equal to mass * gravity:
F_friction = 0.4 * (mass * gravity)
F_friction = 0.4 * (10kg * 9.8m/s^2)
F_friction = 0.4 * 98N
F_friction = 39.2N
3. Force of Gravity: The force of gravity is acting vertically downward on the block. It is given by:
F_gravity = mass * gravity
F_gravity = 10kg * 9.8m/s^2
F_gravity = 98N
Now, let's find the net force acting on the block in the horizontal direction:
Net Force (F_net) = F_horizontal - F_friction
F_net = 59.895N - 39.2N
F_net = 20.695N
Finally, we can use Newton's second law, which states that the net force equals mass times acceleration (F_net = mass * acceleration), to find the acceleration:
acceleration = F_net / mass
acceleration = 20.695N / 10kg
acceleration = 2.0695 m/s^2 (rounded to four decimal places)
Therefore, the acceleration of the block is 2.0695 m/s^2.