Question

The horizontal surface on which the block of mass 5.1 kg slides is frictionless. The force of 62 N acts on the block in a horizontal direction and the force of 124 N acts on the block at an angle of 60 degrees.
What is the magnitude of the resulting acceleration of the block? The acceleration of
gravity is 9.8 m/s^2

Answer 1

Wb = mg = 5.1kg * 9.8N/kg = 49.98 =
Weight of block.

Fb = (49.98N,0deg).

Fp = Fh = 49.98sin(0) = 0 = Force parallel to the plane = Hor. force.

Fv = 49.98cos0 = 49.98N = Force perpendicular to plane.

Fap = (62N,0deg) + (124N,60deg).

X = hor. = 62 + 124cos60 = 124N.
Y = ver. = 124sin60 = 107.39N.

tanA = Y/X = 107.39 / 124 = 0.8660,
A = 40.89 deg.
Fap = X / cosA = 124 / cos40.89 = 164N.
Fap = (164N,40.9deg) = Force applied.

Fn = Fap*cosA = ma,
164cos40.9 = 5.1a,
a = 164cos40.9 / 5.1 = 24.3m/s^2. =
acceleration of block.