Question

A cart loaded with bricks has a total mass of 10.1 kg and is pulled at constant speed by a rope. The rope is inclined at 29.9 ◦ above the horizontal and the cart moves 8 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.7 .

The acceleration of gravity is 9.8 m/s2 .

1. What is the normal force exerted on the cart by the floor?

Answer in units of N.

2. How much work is done on the cart by the rope?

Answer in units of kJ.

3. Note: 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.

Answer 1

1.

Let tension in rope = T

by force balance on cart in vertical direction:

N + T*sin = m*g eq(1)

By force balance in horizontal direction:

T*cos - fr = m*a

here, a = 0 (given, cart moves at constant speed)

fr = friction force acting on cart = *N

then, T*cos - *N = 0

T = *N/cos eq(2)

now, in eq(1),

N + *N*sin/cos = m*g

N(1 + *tan) = m*g

N = normal force exerted by floor on cart = mg/(1 + *tan)

given, m = mass of cart = 10.1 kg

g = 9.81 m/s^2

= 29.9 deg

= coefficient of kinetic friction between ground and cart = 0.7

So, N = 10.1*9.81/(1 + 0.7*tan(29.9 deg))

N = 70.6 N

2.

work done by rope will be:

Wt = T*d*cos

from eq(2),

T = *N/cos

given, d = displacement = 8 m

So, Wt = (*N/cos)*d*cos

Wt = 0.7*70.6*8

Wt = 395.4 J

Wt = 0.395 kJ

3.

work done by friction will be:

Wf = fr*d*cosA

here,

A = angle between friction and displacement = 180 deg

fr = friction force = *N

given, d = displacement = 8 m

So, Wf = (*N)*d*cosA

Wf = 0.7*70.6*8*cos(180 deg)

Wt = -395.4 J

Wt = -0.395 kJ

'here negative signs means energy is lost due to friction.'