Respuesta :
Answer:
W = 0.63 KJ
Explanation:
Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.
W= F*d *cosα Formula (1)
F : force (N)
d : displacement (m)
α : angle between force and displacement
Newton's second law:
∑F = m*a Formula (2)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.
Forces acting on the cart
W: Weight of the cart : In vertical direction
FN : Normal force : perpendicular to the floor
f : Friction force: parallel to the floor
T : tension Force, inclined at θ=24.7° above the horizontal
Calculated of the W
W= m*g
W= 9.13 kg* 9.8 m/s² = 89.47 N
x-y components o the tension force (T)
Tx = Tcosθ = T*cos 24.7° (N)
Ty = Tsin θ = T*sin 24.7° (N)
Calculated of the FN
We apply the formula (2)
∑Fy = m*ay ay = 0
FN +Ty- W = 0
FN = W-Ty
FN = 89.47-T*sin 24.7°
Calculated of the friction force (f)
f = μk*FN
f =(0.597)*( 89.47-T*sin 24.7° )
f= 53.41-0.249T
Calculated of the tension force of the rope (f)
We apply the formula (2) :
∑Fx = m*ax , ax= 0 ,because the speed of the cart is constant
Tx - f = 0
T*cos 24.7°-( 53.41 - 0.249T )= 0
T*cos 24.7° + 0.249T = 53.41
(1.1575)T = 53.41
T= (53.41) / (1.1575)
T= 46.14 N
Work done on the cart by the rope
We apply the formula (1)
W=T*d *cosα
W= (46.14 N)*(15.1 m) *(cos24.7)
W = 632.97 (N*m) = 632.97 (J)
W = 0.63 KJ
The amount of work is required to move the cart 1.5 meters by the rope is 0.63 J.
What is work done?
Work done is the force applied on a body to move it over a distance. Work done for to move a body by the application of force can be given as,
[tex]W=Fd\cos\theta[/tex]
Here, (F) is the magnitude of force and (d) is the distance traveled.
The normal force applied on the body is the difference of weight of the body to the tension force acting on the body in vertical direction. Therefore,
[tex]F_N=mg-F_t\sin(24.7^o)\\F_N=9.13\times9.8-F_t\sin(24.7^o)\\F_N=89.87\times9.8-F_t\sin(24.7^o)[/tex]
The tension force acting on the body in the horizontal direction (downward) is equal to the friction force, (acting upward). Therefore,
[tex]F_t\cos(24.7)=\mu F_N\\F_t\cos(24.7)=0.597(89.87\times9.8-F_t\sin (24.7))\\F_t=46.14[/tex]
Now, find the work done to move the cart 15.1 meters using the above formula as,
[tex]W=(46.14)(1.5)\cos (24.7)\\W=632.97\rm J\\W=0.63\rmkJ[/tex]
Hence, the amount of work is required to move the cart 1.5 meters by the rope is 0.63 J.
Learn more about the work done here;
https://brainly.com/question/25573309
