A 5kg block is being moved to 3m by a force of 50N inclined to the horizontal at an angle of 30°. Assuming that a friction force of 5N opposing its motion ,find the acceleration of the block and the work done by the block.
If the movement was in the direction of the force...
workdone=change in KE + change in PE
and workdone = force*distance=50*3=150Joules
netforce=mass*acceleration
50-mg*sin30=m*A
put mass m, and g in, solve for acceleration A
To find the acceleration of the block, we need to calculate the net force acting on it.
Step 1: Resolve the force into horizontal and vertical components.
The horizontal component of the force can be found using the equation: F_h = F * cos(angle)
F_h = 50N * cos(30°) = 50N * 0.866 = 43.3N
The vertical component of the force can be found using the equation: F_v = F * sin(angle)
F_v = 50N * sin(30°) = 50N * 0.5 = 25N
Step 2: Calculate the net force in the horizontal direction.
The net force in the horizontal direction can be found using the equation:
Net_Force = Force_applied - Force_friction
Net_Force_h = F_h - Force_friction
Net_Force_h = 43.3N - 5N = 38.3N
Step 3: Determine the acceleration.
Since the mass of the block is given as 5 kg, we can use Newton's second law of motion:
F = m * a
38.3N = 5kg * a
Divide both sides of the equation by 5kg:
38.3N / 5kg = a
a = 7.66 m/s^2
Therefore, the acceleration of the block is 7.66 m/s^2.
To find the work done by the block, we can use the equation:
Work = Force_applied * distance
Step 4: Calculate the work done.
The distance the block is being moved is given as 3m, and the applied force is 50N (the magnitude of the force).
Work = 50N * 3m
Work = 150J
Therefore, the work done by the block is 150 Joules.
To find the acceleration of the block, we need to analyze the forces acting on it.
First, let's resolve the force of 50N into its components. The force inclined at an angle of 30° to the horizontal can be split into two components: one parallel to the incline and one perpendicular to the incline.
The component of the force parallel to the incline is given by F_parallel = 50N * sin(30°), which is approximately 25N.
Next, let's calculate the net force acting on the block along the incline. The net force can be calculated by subtracting the friction force opposing motion from the parallel component of the applied force. So, the net force along the incline is given by F_net = F_parallel - Friction = 25N - 5N = 20N.
Using Newton's second law (F = ma), where F is the net force and m is the mass of the block, we can find the acceleration (a) of the block. Rearranging the formula, we have a = F_net / m = 20N / 5kg = 4 m/s².
Therefore, the acceleration of the block is 4 m/s².
Now, let's calculate the work done by the block. Work (W) is given by the formula W = Force * Distance * cos(θ), where θ is the angle between the force and the displacement. In this case, the angle between the force and the displacement is 30°, and the distance is 3m.
The force acting along the incline is 20N (as calculated earlier), so the work done by the block is W = 20N * 3m * cos(30°). Using the cosine of 30° (which is approximately 0.866), we have W = 20N * 3m * 0.866 ≈ 51.96 J (Joules).
Therefore, the work done by the block is approximately 51.96 Joules.