Solving Enigmatic Physics Problems: Acceleration and Frictional Force Examples

[Flashtf]

What is the acceleration of a 20Kg pail of cement that is pulled upward by a rope that applies a force of 208 n to the pail?



A 10kg box is shoved across a floor at an initial speed of 9m/s. It comes to a rest in 3 seconds after it is shoved. Find the frictional force. Problem begins after the shove is released.

Answer in reply section making sure that answer are identifiable as different..

 

[nrqed]

What is the acceleration of a 20Kg pail of cement that is pulled upward by a rope that applies a force of 208 n to the pail?



A 10kg box is shoved across a floor at an initial speed of 9m/s. It comes to a rest in 3 seconds after it is shoved. Find the frictional force. Problem begins after the shove is released.

Answer in reply section making sure that answer are identifiable as different..

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[ogb p]

The acceleration of the 20kg pail of cement can be calculated using the formula F=ma, where F is the force applied, m is the mass of the object, and a is the acceleration. In this case, the force applied is 208N, and the mass is 20kg. Therefore, the acceleration of the pail is 208N/20kg = 10.4 m/s^2.

For the 10kg box, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 9m/s, the final velocity is 0m/s (since it comes to a rest), and the time is 3 seconds. Therefore, the acceleration of the box is -3m/s^2 (negative sign indicates deceleration).

To find the frictional force, we can use the formula F = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the box, which is mg, where g is the acceleration due to gravity (9.8m/s^2). So, N = 10kg x 9.8m/s^2 = 98N.

Now, we need to find the coefficient of friction. We can use the formula μ = F/N. In this case, the force acting on the box is the force of friction, which is equal to the mass of the box multiplied by its acceleration (F=ma). So, μ = (10kg)(-3m/s^2)/98N = -0.306.

Therefore, the frictional force acting on the box is μN = (-0.306)(98N) = -30N. This means that the frictional force acting on the box is 30N in the opposite direction of its motion.