A block of mass m is pushed along a horizontal table by a force F= 70 N at an angle 55 degrees with the horizontal. The coefficient of kinetic friction is 0.3
To find the net force acting on the block, we need to break down the force F into its horizontal and vertical components.
The horizontal component of the force, Fx, can be found using trigonometry:
Fx = F * cos(theta)
where F is the magnitude of the force (70 N) and theta is the angle with the horizontal (55 degrees).
Substituting the given values:
Fx = 70 N * cos(55 degrees)
Fx ≈ 70 N * 0.5736
Fx ≈ 40 N
The vertical component of the force, Fy, can also be determined using trigonometry:
Fy = F * sin(theta)
Substituting the values:
Fy = 70 N * sin(55 degrees)
Fy ≈ 70 N * 0.8192
Fy ≈ 57.34 N
Now, we can calculate the frictional force, Ff, using the equation:
Ff = coefficient of friction * Normal force
The normal force, Fn, is the force exerted by the table perpendicular to the block's surface. In this case, since the block is on a horizontal table, Fn is equal to the block's weight:
Fn = m * g
where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the coefficient of kinetic friction (0.3):
Ff = 0.3 * (m * g)
We can now determine the net force acting on the block in the horizontal direction:
F_net = Fx - Ff
Substituting the previously calculated values:
F_net = 40 N - (0.3 * (m * 9.8 m/s^2))
F_net = 40 N - (2.94 m * 9.8 m/s^2)
F_net = 40 N - (28.692 m)
F_net = 40 N - 28.692 m
Thus, the net force acting on the block is (40 N - 28.692 m) in the horizontal direction.