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
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
2 answers
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
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
2 answers