A cart loaded with bricks has a total mass of 23.5 kg and is pulled at constant

Question

A cart loaded with bricks has a total mass
of 23.5 kg and is pulled at constant speed by
a rope. The rope is inclined at 23.4? above
the horizontal and the cart moves 18.1 m on
a horizontal ?oor. The coe?cient of kinetic
friction between ground and cart is 0.8 .
The acceleration of gravity is 9.8 m/s^2. How much work is done on the cart by the
rope?
Answer in units of kJ

This is the work I have done so far....
equation 1:
x axis: F cos @ - fk = 0 -----> F cos @ = Uk (mg + F sin @)
equation 2:
y axis: Fn + F sin @ - mg = 0 -------> F cos @ -Uk F sin @
equation 3:
fk = uk*Fn --------> F(cos @ - Uk sin @) = Uk mg

Fk = UFn (let us call the Eqn A
Ft cos @ = fk (Eqn B)
Ft cos @ = UFn (Eqn C)

Ft = U Fn/cos @

Fn = (Ft cos @)/u (Eqn D)

f cos @ -Uk*Fn = 0
F cos @ = Uk*Fn
Fn = (F cos @)/Uk .........now plug into eqn 2

(Ft cos @)/Uk + F sin @ -mg = 0

F (cos @/Uk + sin @) = mg

solve for f
f(cos @ + sin @) = mgUk (Eqn E)
f = 23.5*9.8*.8/cos 23 + sin 23
f = 184.24 ...............this was incorrect in Quest, not sure where I went wrong

Explanation / Answer

x-direction --> Fcos@ = u*(mg - Fsin@) F = umg/(cos@ + u*sin@) = 149.13 N Work done = Fcos@ * s = 149.13cos23.4 * 18.1 = 2.477 kJ

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