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a) See attached
b) The block's weight has magnitude
mg = (25 kg) (9.8 m/s²) = 245 N
and we split this vector into components acting parallel and perpendicular to the incline, with magnitudes
• parallel : mg sin(40°) ≈ 157 N
• perpendicular : mg cos(40°) ≈ 188 N
Take "down the plane" to be the positive parallel direction, and the direction of the normal force to be the positive perpendicular direction. Then these components of the weight are positive and negative,
• parallel : mg sin(40°) ≈ 157 N
• perpendicular : mg cos(40°) ≈ -188 N
c) The net force acting parallel to the incline is
∑ F[para] = mg sin(40°) - f = 0
where f is the magnitude of kinetic friction. The net force is zero since the block slides at a constant speed.
d) Solve the equation in (c) for f :
f = mg sin(40°) ≈ 157 N
e) The magnitude of kinetic friction f is proportional to the magnitude of the normal force n by a factor of µ, the coefficient of friction.
f = µn
The net force on the block acting perpendicular to the incline is
∑ F[perp] = n - mg cos(40°) = 0
and it follows that
n = mg cos(40°) ≈ 188 N
Solve for µ :
157 N = µ (188 N) ⇒ µ ≈ 0.84
