Question
. A body of mass 5kg is initially at rest at the bottom of a rough inclined plane of length 6.3m. The plane is inclined at 300 to the horizontal and coefficient of friction between the body and the plane is 1 2√ . A constant horizontal force of 35 √ 3N is applied to the body causing it to 3 accelerate up the plane. Find the time taken for the body to reach the top and its speed on arrival.
Answers
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To solve this problem, we need to use the equations of motion and the principles of friction. Let's break it down step by step:
1. Draw a diagram:
2. Determine the forces acting on the body:
- Weight of the body (mg) = 5 x 9.81 = 49.05 N acting vertically downwards
- Normal force (N) = mg cos(30°) = 49.05 x cos(30°) = 42.5 N acting perpendicularly to the inclined plane
- Force applied (F) = 35√3 N acting horizontally up the plane
- Frictional force (f) = N x coefficient of friction = 42.5 x 1/2√ = 21.25√2 N acting parallel to the inclined plane in the opposite direction to the direction of motion
3. Determine the net force acting on the body:
- Forces acting up the plane = F sin(30°) - f = 35√3 x sin(30°) - 21.25√2 = 22.5 N
- Forces acting down the plane = mg sin(30°) - N - F cos(30°) = 5 x 9.81 x sin(30°) - 42.5 - 35√3 x cos(30°) = -15.86 N
- Net force acting on the body = forces acting up the plane + forces acting down the plane = 22.5 - 15.86 = 6.64 N acting up the plane
4. Determine the acceleration of the body:
- Using Newton's second law (F = ma), we can determine the acceleration of the body.
- Net force acting on the body = ma
- Therefore, acceleration (a) = net force/mass = 6.64/5 = 1.328 m/s^2
5. Determine the time taken for the body to reach the top:
- Using the equation of motion s = ut + 1/2at^2, where s is the distance travelled, u is the initial velocity (which is zero), a is the acceleration, and t is the time taken
- Distance travelled up the plane (s) = length of inclined plane (6.3 m)
- Therefore, 6.3 = 1/2 x 1.328 x t^2
- Solving for t, we get t = 2.63 s (rounded to two decimal places)
6. Determine the speed of the body on arrival:
- Using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration, and t is the time taken
- Final velocity (v) = u + at = 0 + 1.328 x 2.63 = 3.49 m/s (rounded to two decimal places)
Therefore, the time taken for the body to reach the top is 2.63 seconds and the speed of the body on arrival is 3.49 m/s.
1. Draw a diagram:
2. Determine the forces acting on the body:
- Weight of the body (mg) = 5 x 9.81 = 49.05 N acting vertically downwards
- Normal force (N) = mg cos(30°) = 49.05 x cos(30°) = 42.5 N acting perpendicularly to the inclined plane
- Force applied (F) = 35√3 N acting horizontally up the plane
- Frictional force (f) = N x coefficient of friction = 42.5 x 1/2√ = 21.25√2 N acting parallel to the inclined plane in the opposite direction to the direction of motion
3. Determine the net force acting on the body:
- Forces acting up the plane = F sin(30°) - f = 35√3 x sin(30°) - 21.25√2 = 22.5 N
- Forces acting down the plane = mg sin(30°) - N - F cos(30°) = 5 x 9.81 x sin(30°) - 42.5 - 35√3 x cos(30°) = -15.86 N
- Net force acting on the body = forces acting up the plane + forces acting down the plane = 22.5 - 15.86 = 6.64 N acting up the plane
4. Determine the acceleration of the body:
- Using Newton's second law (F = ma), we can determine the acceleration of the body.
- Net force acting on the body = ma
- Therefore, acceleration (a) = net force/mass = 6.64/5 = 1.328 m/s^2
5. Determine the time taken for the body to reach the top:
- Using the equation of motion s = ut + 1/2at^2, where s is the distance travelled, u is the initial velocity (which is zero), a is the acceleration, and t is the time taken
- Distance travelled up the plane (s) = length of inclined plane (6.3 m)
- Therefore, 6.3 = 1/2 x 1.328 x t^2
- Solving for t, we get t = 2.63 s (rounded to two decimal places)
6. Determine the speed of the body on arrival:
- Using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity (which is zero), a is the acceleration, and t is the time taken
- Final velocity (v) = u + at = 0 + 1.328 x 2.63 = 3.49 m/s (rounded to two decimal places)
Therefore, the time taken for the body to reach the top is 2.63 seconds and the speed of the body on arrival is 3.49 m/s.