Posted by **Kylie** on Saturday, October 1, 2011 at 8:29am.

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

- physics -
**Henry**, Monday, October 3, 2011 at 3:21pm
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.

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