Answer:
Fapp = 3 N
Explanation:
[tex]W_{net} = F_{app} * d (1)[/tex]
[tex]\Delta K = K_{f} - K_{i} = \frac{1}2}* m* (v_{f} ^{2} -v_{i} ^{2} ) (2)[/tex]
[tex]F_{app} =(\frac{1}2}* m* (v_{f} ^{2} -v_{i} ^{2} ) ) / d = (5 kg* (2m/s)^{2} -1m/s^{2})/ 5 m = 3 N[/tex]
With the use of Newton's second law formula, the applied force is 3 Newtons
Given that a 10 kg box slides horizontally without friction at a speed of 1 m/s. At one point, a constant force is applied to the box in the direction of its motion. The box travels 5 m with the constant force applied. The force is then removed, leaving the box with a speed of 2 m/s.
Initial velocity U = 1 m/s
Final velocity V = 2 m/s
mass m = 10 kg
Distance s = 5 m
Applied force F = ?
Let us first calculate the acceleration of the box by using third equation of motion.
[tex]V^{2} = U^{2} + 2aS[/tex]
Substitute all the parameters
[tex]2^{2} = 1^{2}[/tex] + 2 x a x 5
4 = 1 + 10a
10a = 3
a = 3/10
a = 0.3 m/[tex]s^{2}[/tex]
To calculate the magnitude of the applied force, use Newton's second law formula. That is,
F = ma
Substitute all the necessary parameters
F = 10 x 0.3
F = 3N
Therefore, the applied force is 3 Newtons
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