A block of mass 7.5 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of 31.0 N at an angle 26.3° above the horizontal. What is the magnitude of the acceleration of the block?
I have nooo clue what to do for this one. Can someone please show me the steps?
Well, the component of force horiontal, doing the pulling is 31cosTheta.
Forcehorizontal= mass* acceleration
solve for acceleration.
Now it asks for the normal force exerted on the mass by the floor. Do I need to use the acceleration for that?
To find the magnitude of the acceleration of the block, we can break down the given force into its horizontal and vertical components.
1. Resolve the force into its vertical and horizontal components:
Given force = 31.0 N at an angle of 26.3° above the horizontal.
Vertical component: Fv = F × sin θ
Fv = 31.0 N × sin(26.3°)
Horizontal component: Fh = F × cos θ
Fh = 31.0 N × cos(26.3°)
2. Now, let's apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
∑F = m × a
3. Since the horizontal surface is frictionless, the only force acting on the block is the horizontal component of the applied force (Fh = 31.0 N × cos(26.3°)).
Therefore, the net force (Fnet) is equal to the horizontal component of the applied force (Fh = 31.0 N × cos(26.3°)).
Fnet = Fh = 31.0 N × cos(26.3°)
4. Substitute the values into Newton's second law and solve for acceleration:
Fnet = m × a
31.0 N × cos(26.3°) = 7.5 kg × a
To find the magnitude of the acceleration (a), divide both sides by 7.5 kg:
a = (31.0 N × cos(26.3°)) / 7.5 kg
5. Calculate the value of acceleration using a calculator:
a = (31.0 N × cos(26.3°)) / 7.5 kg
The calculated value will give you the magnitude of the acceleration of the block.