The horizontal surface on which the block of mass 5.1 kg slides is frictionless. The force of 62 N acts on the block in a horizontal direction and the force of 124 N acts on the block at an angle of 60 degrees.

What is the magnitude of the resulting acceleration of the block? The acceleration of
gravity is 9.8 m/s^2

Well, let's see how fast this block is going to get its groove on!

First, we need to break down the force acting on the block. We got a force of 62 N in a straight line, and another force of 124 N at an angle of 60 degrees. So, if we do some trigonometry magic, we can find that the force acting in the horizontal direction is 62 N + cos(60°) x 124 N = 62 N + 62 N = 124 N.

Now, let's apply Newton's second law, F = ma. The net force acting on the block is 124 N, and the mass is 5.1 kg. So, we can rearrange the equation to solve for the acceleration: a = F/m = 124 N / 5.1 kg = 24.3 m/s^2.

So, the magnitude of the resulting acceleration is 24.3 m/s^2. That block is going to speed things up faster than a clown on roller skates!

To find the magnitude of the resulting acceleration of the block, we can use Newton's second law of motion:

Fnet = ma

Where:
Fnet is the net force acting on the block
m is the mass of the block
a is the resulting acceleration of the block

First, we need to calculate the net force acting on the block. The force of 62 N acts horizontally, and the force of 124 N acts at an angle of 60 degrees. We can break down the force into horizontal and vertical components.

The horizontal component of the force at an angle of 60 degrees can be calculated using trigonometry:

Fhorizontal = F * cos(θ)
= 124 N * cos(60°)
= 124 N * 0.5
= 62 N

Now we can calculate the net force:

Fnet = Fhorizontal + Fhorizontal
= 62 N + 62 N
= 124 N

Next, we can use Newton's second law to find the resulting acceleration:

Fnet = ma
124 N = 5.1 kg * a

To find the magnitude of the resulting acceleration, divide both sides of the equation by the mass:

a = 124 N / 5.1 kg
≈ 24.3 m/s^2

So, the magnitude of the resulting acceleration of the block is approximately 24.3 m/s^2.

To find the magnitude of the resulting acceleration of the block, we need to use Newton's second law of motion. Newton's second law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration.

In this case, we have two forces acting on the block: the horizontal force of 62 N and the force of 124 N at an angle of 60 degrees. The net force is the vector sum of these two forces. We can find the horizontal and vertical components of the force at an angle of 60 degrees using trigonometry.

The horizontal component (Fh) of the force is given by:
Fh = 124 N * cos(60°)

The vertical component (Fv) of the force is given by:
Fv = 124 N * sin(60°)

Since the horizontal surface is frictionless, the force of 62 N is the only force that will cause the block to accelerate horizontally. Therefore, the net force in the horizontal direction is equal to Fh - 62 N.

Now, we can calculate the magnitude of acceleration (a) using Newton's second law:

F_net = m * a

Where F_net is the net force in the horizontal direction and m is the mass of the block.

First, we need to calculate the net force in the horizontal direction:

F_net = Fh - 62 N

Next, substitute the given values into the formula:

F_net = (124 N * cos(60°)) - 62 N

Now, we can calculate the magnitude of acceleration:

a = F_net / m

Substitute the values of F_net and m into the formula:

a = ((124 N * cos(60°)) - 62 N) / 5.1 kg

Finally, calculate the result:

a ≈ [((124 N * cos(60°)) - 62 N) / 5.1 kg] m/s^2

Please note that since we do not know the exact value of cos(60°), you may need to use a calculator or a trigonometric table to find the cosine of 60 degrees.

Wb = mg = 5.1kg * 9.8N/kg = 49.98 =

Weight of block.

Fb = (49.98N,0deg).

Fp = Fh = 49.98sin(0) = 0 = Force parallel to the plane = Hor. force.

Fv = 49.98cos0 = 49.98N = Force perpendicular to plane.

Fap = (62N,0deg) + (124N,60deg).

X = hor. = 62 + 124cos60 = 124N.
Y = ver. = 124sin60 = 107.39N.

tanA = Y/X = 107.39 / 124 = 0.8660,
A = 40.89 deg.
Fap = X / cosA = 124 / cos40.89 = 164N.
Fap = (164N,40.9deg) = Force applied.

Fn = Fap*cosA = ma,
164cos40.9 = 5.1a,
a = 164cos40.9 / 5.1 = 24.3m/s^2. =
acceleration of block.