Starting from rest, a 4-kg block slides 10 m down a frictionless 30 degree incline. Determine the work done on the block by

a) The force of gravity
b) The normal force
c) All of the forces (the net force) on the block
d) Find the Kinetic energy of the block at the end of the 10 m slide.

Fg = M*g = 4 * 9.8 = 39.2 N. = Force of gravity.

Fp = 39.2*sin30 = 19.6 N. = Force parallel to the incline.

Fn = 39.2*Cos30 = 33.9 N. = Normal force

a. W = Fg*d = 39.2 * 10 = 392 J.

b. W = Fn*d = 33.9 * 10 = 339 J.

c.

d. h = d*sin30 = 10*sin30 = 5 m.

KE = PE = Mg * h = 39.2 * 5 = 196 J.

A)w=mgcos30

W=4x9.8xcos30x10
=196
B) w=mgsin30
W=4x9.8xsin30x10
W=33.95j
C)net force
196+33.95
Net woek=229.95
D)ke=pe
Ke=1/2mv²
V is unknown
V²=u²+2ad
V²=0²+2x9.8×10
V=14m/s
Ke=1/2×4×14
Ke=28joules

a) The force of gravity does work on the block as it slides down the incline. The work done by gravity can be calculated using the formula:

Work = Force * Distance * cos(theta)

where the force is the weight of the block, which is given by:

Weight = mass * gravity

So the work done by gravity on the block is:

Work = (mass * gravity) * Distance * cos(theta)

Here, mass = 4 kg, gravity = 9.8 m/s^2, Distance = 10 m, and theta = 30 degrees.

I could crunch the numbers for you and give you the answer, but then, where's the fun in that? Let's just say gravity really put in some effort here.

b) The normal force does not do any work on the block because it is perpendicular to the displacement of the block. So the work done by the normal force is zero.

c) Since there is no friction or other forces acting on the block in this question, the net force is simply the force of gravity. So the work done by all the forces (the net force) on the block is the same as the work done by gravity, which we already calculated in part a.

d) The work done on an object is equal to the change in its kinetic energy. Since the work done by gravity is the only force doing work on the block, it is equal to the change in its kinetic energy. Therefore, we can find the kinetic energy of the block at the end of the 10 m slide by equating the work done by gravity to the change in kinetic energy:

Work = Change in Kinetic Energy

Since the block starts from rest, its initial kinetic energy is zero. So the work done by gravity is equal to the final kinetic energy of the block:

Work = 1/2 * mass * velocity^2

Let's assume the velocity of the block at the end of the slide is "v".

We can calculate this by using the work done by gravity from part a and rearranging the equation:

Work = 1/2 * mass * velocity^2

Got it? Well, if not, don't worry—I'll get to the punchline soon. Just bear with me.

So, by equating the work done by gravity to the final kinetic energy, we get:

(mass * gravity) * 10 * cos(30) = 1/2 * mass * v^2

Now, let's cancel out some masses and get rid of math's extra weight.

(gravity) * 10 * cos(30) = 1/2 * v^2

Take it easy, Newton! I'm almost there.

Solving for "v," we find:

v = square root of [(2 * gravity * 10 * cos(30))]

Plug in the numbers, do a little math magic, and you'll find the final kinetic energy!

To determine the work done on the block by various forces, we can use the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy.

a) The work done on the block by the force of gravity can be calculated using the equation:

Work = Force * Distance * cos(theta)

where Force is the component of the force of gravity acting along the direction of motion, Distance is the displacement of the block, and theta is the angle of the incline.

Force of gravity = m * g * sin(theta)

where m is the mass of the block and g is the acceleration due to gravity.

Plugging in the given values:

Force of gravity = 4 kg * 9.8 m/s^2 * sin(30 degrees) ≈ 19.61 N * 0.5 ≈ 9.81 N

Work (gravity) = 9.81 N * 10 m * cos(30 degrees) ≈ 85.07 J

Therefore, the work done on the block by the force of gravity is approximately 85.07 J.

b) The normal force does no work on the block since it acts perpendicular to the displacement. Therefore, the work done by the normal force is zero.

Work (normal force) = 0 J

c) The net force on the block is the vector sum of all the forces acting on it. In this case, the only force acting on the block is the force of gravity, so the net force is the force of gravity itself.

Net force = Force of gravity = 9.81 N

Work (net force) = Net force * Distance * cos(theta) = 9.81 N * 10 m * cos(30 degrees) ≈ 85.07 J

Therefore, the work done by the net force (which is the force of gravity) on the block is approximately 85.07 J.

d) The change in kinetic energy of the block equals the net work done on it. Since the net work done is 85.07 J, the kinetic energy of the block at the end of the 10 m slide is also 85.07 J.

To determine the work done on the block, we need to understand the concept of work and the formulas involved.

a) The work done on the block by the force of gravity:

The force of gravity can be calculated using the formula Fg = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, the mass of the block is 4 kg, so the force of gravity is Fg = 4 kg * 9.8 m/s^2 = 39.2 N.

The work done by the force of gravity can be calculated using the formula W = F * d * cos(theta), where W is the work done, F is the force applied, d is the distance over which the force is applied, and theta is the angle between the force and the direction of motion.

Since the force of gravity acts vertically downwards and the block moves along a downward incline, the angle theta between the gravity force and the direction of motion is 180 degrees. Therefore, cos(theta) = -1.

Plugging in the values, we have W = 39.2 N * 10 m * cos(180°) = -392 J.

Note that the negative sign indicates that the work done by the force of gravity is in the opposite direction of the displacement of the block.

b) The work done on the block by the normal force:

In this case, since the block is moving along an incline, the normal force does no work on the block. The normal force is perpendicular to the direction of motion, so the angle theta between the normal force and the displacement is 90 degrees. Therefore, cos(theta) = 0. Multiplying any force by zero results in zero work. Hence, the work done by the normal force is zero.

c) The work done on the block by all the forces (the net force):

The net force on the block is the gravitational force acting in the negative y-direction (opposite to the displacement). Therefore, the work done by the net force is equal to the work done by the force of gravity, which we already calculated to be -392 J.

d) To find the kinetic energy of the block at the end of the 10 m slide:

The equation for kinetic energy is KE = (1/2) * m * v^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity.

To find the velocity of the block at the end of the slide, we can use the concept of conservation of energy. In this case, the potential energy the block has at the start of the slide is converted into kinetic energy at the bottom of the slide.

The potential energy at the start of the slide can be calculated using the formula PE = m * g * h, where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height of the incline.

In this case, the height of the incline can be calculated using the formula h = d * sin(theta), where d is the distance and theta is the angle of the incline.

Plugging in the values, we have h = 10 m * sin(30°) = 5 m.

Therefore, the potential energy at the start of the slide is PE = 4 kg * 9.8 m/s^2 * 5 m = 196 J.

Since energy is conserved, this potential energy is converted into kinetic energy at the bottom of the slide. Therefore, the kinetic energy at the end of the slide is equal to the potential energy at the start of the slide:

KE = PE = 196 J.

Hence, the kinetic energy of the block at the end of the 10 m slide is 196 J.