Question

A cart loaded with bricks has a total mass of 17 kg and is pulled at constant speed by a rope. The rope is inclined at 25 degrees above the horizontal and the cart moves 22.8 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.6. The acceleration of gravity is 9.8 m/s^2.a. What is the normal force exerted on the cart by the floor? Answer in units of N.b. How much work is done on the cart by the rope? Answer in kJc. Note: The energy change due to friction is a loss of energy. What is the energy change Wf due to friction ? Answer in kJ.

Answer 1

To find the normal force exerted on the cart by the floor, we need to consider the forces acting on the cart. The normal force is equal to the sum of the gravitational force and the force due to the incline.

Step-by-step explanation:

To find the normal force exerted on the cart by the floor, we need to consider the forces acting on the cart. The gravitational force (mg) acts vertically downward, and the force due to the incline (mg*sin(theta)) acts perpendicular to the incline. Since the cart is moving at a constant speed, the net force is zero, so the normal force is equal to the sum of these two forces. The normal force is given by:

Fn = mg + mg*sin(theta)

Given that the mass of the cart (m) is 17 kg, the acceleration due to gravity (g) is 9.8 m/s^2, and the angle (theta) is 25 degrees, we can substitute these values into the formula to find the normal force:

Fn = (17 kg)(9.8 m/s²) + (17 kg)(9.8 m/s²)*sin(25 degrees)

Fn ≈ 174.34 + 73.12 ≈ 247.46 N

Answer 2

a) N = 382.9 N

b) W = 5.237 kJ

c) ΔE = Wf =- 5.237 kJ

Step-by-step explanation:

Newton's second law :

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=17 kg : mass of the cart

θ = 25°: angle of the Tension force above the horizontal

μk= 0.60: coefficient of kinetic friction between the cart and the ground

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the cart

We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.

W: Weight of the cart : In vertical direction downaward

N : Normal force : In vertical direction the upaward

T : Force applied to the cart

f : Friction force: In horizontal direction

Calculated of the weight of the cart

W= m*g = (17 kg)*(9.8 m/s²)= 166.6 N

x-y components T

Tx = Tcosθ = T*cos(25)°

Ty = Tsinθ = T*sin(25)°

Calculated of the Normal force

∑Fy = m*ay ay = 0

N-W+Ty= 0

N-W+T*sin(25)= 0

N = 490 -T*sin(25)° Equation (1)

Calculated of the kinetic friction force (fk):

fk=μk*N= 0.6*( 490 -T*sin(25)°

fk = 0.6*490 -0.6T*sin(25)°

fk = 294 -0.6¨*sin(25)° T Equation (2)

Newton's second law to the cart

∑F = m*a

a =0 because the cart moves at constant speed

∑F = 0

Tx-fk=0

T*cos(25)°= fk Equation (3)

a) Calculated of the Normal force

Equation (2) = Equation (3) = fk

294 -0.6¨*sin(25)° T=T*cos(25)°

294 =T(0.6¨*sin(25)°)+ T(cos(25)°)

294 =T(0.6¨*sin(25)° + cos(25)°)

294 =T(0.6¨*sin(25)°+(cos(25)°)

294 =T(0.6¨*sin(25)°+(cos(25)°)

294 = T(1.16)

T = (294) /(1.16)

T = 253.45 N

We replace T = 253.45 N in the equation (1)

N = 490 -T*sin(25)°

N = 490 - ( 253.45)*sin(25)°

N = 382.9 N

b) Work done by the rope on the cart

W = (Tx) *d

W = (T*cos(25)°)*(22.8)

W = (253.45*cos(25)°)(22.8) (N*m)

W = 5237.24 (N*m)

W = 5237.24 J = 5.237 kJ

What is the energy change Wf due to friction ?

ΔE= Wfk

ΔE : Energy change

Wfk : Work done by fk

in the Equation (3) :

T*cos(25)°= fk = ( 253.45) N*cos(25)° = 229.7 N

ΔE= - fk*d

ΔE= - (229.7 N)*(22.8 )m

ΔE= - 5.237 kJ =Wfk