A cart loaded with bricks has a total mass of 25.2 kg and is pulled at constant speed by a rope. The rope is inclined at 21.6 ◦ above the horizontal and the cart moves 13.7 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.4 .

The acceleration of gravity is 9.8 m/s2 .
What is the normal force exerted on the cart by the floor? Answer in units of N.

How much work is done on the cart by the rope?
Answer in units of kJ.

The energy change due to friction is a loss of energy.
What is the energy change Wf due to fric- tion?
Answer in units of kJ.

6 answers

(a)
Summing the components of the forces along the vertical direction provides
F• sinα + N - m•g = 0,
=> F• sinα = m•g - N.
For the horizontal direction we have
F•cosα – F(fr) = 0
F(fr) = k•N, => F•cosα = k•N,
F• sinα/ F•cosα = (m•g – N)/k•N,
tan α = (m•g – N)/k•N,
N = m•g/(k•tan +1) = 25.2•9.8/(0.4•tan21.6º +1) = 213 N.
(b)
If F•cosα = k•N,
F = k•N/cosα.
The work done on the cart is given by
W(fr) =F•s•cosα = k•N•s•cos α/cosα = =k•N•s = 0.4•213•13.7 = 1167 J =
=1.167 kJ.
(c)
The energy lost due to friction is just the work done by friction on the cart:
W(fr) = F(fr)•s•cos 180º = k•N•s•(-1) =
= - k•N•s = - 0.4•213•13.7 =
= - 1167 J = - 1.167 kJ.

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