A box with weight 55 N is on a rough horizontal surface. An external force F is applied horizontally to the box. A normal force and a friction force are also present, denoted by n and f.A force diagram, showing the four forces that act on the box, is shown below:When force F equals 2.0 N, the box is in motion at constant velocity.What is the acceleration of the box when force F equals 2.6 N?Express your answer in m/s2.
Question
A box with weight 55 N is on a rough horizontal surface. An external force F is applied horizontally to the box. A normal force and a friction force are also present, denoted by n and f.A force diagram, showing the four forces that act on the box, is shown below:When force F equals 2.0 N, the box is in motion at constant velocity.What is the acceleration of the box when force F equals 2.6 N?Express your answer in m/s2.
Solution
To solve this problem, we need to understand that when an object is moving at a constant velocity, the net force acting on it is zero. This means that the applied force (F) is equal to the frictional force (f).
From the problem, we know that when F = 2.0 N, the box is moving at a constant velocity. Therefore, the frictional force (f) is also 2.0 N.
Now, when the force F is increased to 2.6 N, the net force acting on the box is F - f = 2.6 N - 2.0 N = 0.6 N.
The acceleration (a) of the box can be found using Newton's second law, F = ma. Rearranging for acceleration gives a = F/m.
However, we don't have the mass of the box. We can find it using the equation for weight, W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. Rearranging for mass gives m = W/g.
Substituting the given weight of the box (55 N) and the acceleration due to gravity (9.8 m/s²) gives m = 55 N / 9.8 m/s² = 5.61 kg.
Finally, substituting the net force (0.6 N) and the mass (5.61 kg) into the equation for acceleration gives a = 0.6 N / 5.61 kg = 0.107 m/s².
So, the acceleration of the box when the force F equals 2.6 N is approximately 0.107 m/s².
Similar Questions
This image shows a box on the ground, the normal force acting on the box, and the force being applied to the box. The box is accelerating under the applied force. If the kinetic friction coefficient between the box and the ground is 0.60, what is the friction force acting on the box?Select one:7.40 N to the right.highlight_off9.00 N to the right.highlight_off32.0 N to the right.highlight_off24.6 N to the right.highlight_off
A box with mass 4.0 kg is pulled from rest along a horizontal floor by a constant force F. A constant frictional force of 1.0 N acts on the box. When the box has travelled 1.0 m its velocity is 1.5 m / s.The kinetic energy of the box is equal to the work done on the box.What is force F ?
A small box of mass m is placed on a top of a larger box of mass 2m as shown in the diagram. When a force F is applied to the large box, both boxes accelerate to the right with the same acceleration. If the coefficient of friction between all surfaces is μ what would be the force accelerating the smaller mass?
A pulling force of 7 N to the left , a pushing force of 4 N to the right and a pulling force of 4 N to the right acts on a box that is on a frictionless surface. What is the net force acting on the box?
8. A 50 kg box was slid along a floor by an applied force of 200 N. If the coefficient of frictionwas 0.50, what is its acceleration?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.