Answer:
Explanation:
Let force acting on the card = F . let its x component be F_x
F_x . 3 = 300
F_x = 100
similarly ,
F_y . 4 = 400
F_y = 100
Total F = √ 3² + 4²
= 5 N
2 ) Force of the cart on the student = 5 N ( Reaction force)
3) Distance moved by the cart = 3 + 4 = 7m
4) magnitude of displacement = 5 m
The force that is exerted by the student on the cart is 141.4 N.
The force in the horizontal direction is obtained from;
Work = Force × distance
Force = Work/distance = 300 J/3 m = 100 N
The force in the vertical direction;
Work = Force × distance
Force = Work/distance = 400 J/ 4 m = 100 N
The resultant force is the force that student exerts on the cart which is;
R = √(100)^2 + (100)^2
R = 141.4 N
The cart exerts the same magnitude of force on the boy but in opposite direction. Hence, force of the cart on the boy = -141.4 N.
The total distance is obtained by an algebraic sum since distance is not a vector; 3 m + 4 m= 7 m
The displacement is obtained from;
D = √(3)^2 + (4)^2
D = 5 m
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