physics
posted by phys1! .
A cart loaded with bricks has a total mass
of 18 kg and is pulled at constant speed by
a rope. The rope is inclined at 30.4 above
the horizontal and the cart moves 21.5 m on
a horizontal floor. The coefficient of kinetic
friction between ground and cart is 0.8 .
The acceleration of gravity is 9.8 m/s2 .
How much work is done on the cart by the
rope?
Answer in units of kJ

Calculate the rope tension from the information given.
Call it T.
The horizontal component of T = Friction force
since it does not accelerate.
(M*g  T sin21.5) *0.8 = T cos 21.5
Solve for T
Work done by rope = T*cos30.4*21.5 m 
T*cos(θ) = Ffr –––> Ffr = µk* Fn –––>
Fn = m*g  T*sin(θ) :
T*cos(θ) = ( µk*(m*g  T*sin(θ) )
T*cos(30.4º) = (0.8 *(18*9.8 – T*sin(30.4º))
T*cos(30.4º) = 88.2 N – 0.8 * T *sin(30.4º)
T*cos(30.4º) + 0.8*T*sin(30.4º) = 141.12 N
T (cos(30.4º) + 0.8*sin(30.4º)) = 141.12 N
T = 141.12 N / (cos(30.4º) + 0.8*sin(30.4º))
T = 111.4 N

Work = T*cos(θ)*d
W = 111.4*cos(30.4º)*21.5m = 2065.81 J or 2.07 kJ