The force exerted by student A with his scooter is 306 N and that of student B is 204 N.
<h3>
Force applied by each student</h3>
The force exerted by each student is calculated from Newton's second law of motion.
F = ma
where;
- m is mass
- a is acceleration
F(A) = 127.5 x 2.4
F(A) = 306 N
F(B) = 120 x 1.7
F(B) = 204 N
Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.
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d. natural gas is the answer
Answer:
R=0.5B+0.5C+2A+D
Explanation:
By the triangular law of vector addition
vector R= vector B- vector D
As A,B,C,D are edges of the parallelogram,
A is parallel to D but opposite in direction.
Therefore
;
;

B is parallel to C and in same direction.



The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force

acting against the motion. Since their resultant must be zero, we have:

The frictional force is

where

is the coefficient of kinetic friction

is the weight of the block.
Substituting these values, we find the magnitude of the force F: