41kg object that is moving east at 5 m s
Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as


Now the normal force on the block is given as



now the friction force on the cart is given as



now if cart moves with constant speed then net force on cart must be zero
so now we have




so the force must be 199.2 N
Answer:
The strategy we would like you to learn has five major steps: Focus the Problem, Physics Description, Plan a Solution, Execute the Plan, and Evaluate the Solution. Let's take a detailed look at each of these steps and then do an sample problem following the strategy.
Answer:
Approximately
, assuming that the volume of these two charged objects is negligible.
Explanation:
Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.
Let
and
denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question,
and
.
Let
denote the distance between these two point charges. In this question,
.
Let
denote the Coulomb constant. In standard units,
.
By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:
.
Substitute in the values and evaluate:
.
You are correct...amplitude will be the answer