contestada

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.

Respuesta :

valeriagonzalez0213

Answer:

a) N = 382.9 N

b) W = 5.237 kJ

c) ΔE = Wf =- 5.237 kJ

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