What is the net work on the block?

A 5 kg block is released from rest at the top of a quarter- circle type curved frictionless surface. The radius of the curvature

malfutka [58]

Answer:

a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J

Explanation:

a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.

So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.

So, mgh + 0 = 0 + K'

K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m

So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J

b. Since the kinetic energy of the block K = 1/2mv² where m = mass of block = 5 kg, v = velocity of block at bottom of curve

So, v = √(2K/m)

= √(2 × 186.2 J/5 kg)

= √(372.4 J/5 kg)

= √(74.48 J/kg)

= 8.63 m/s

c. To find the stopping distance, from work-kinetic energy principles,

work done by friction = kinetic energy change of block.

So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance

ΔK = -fd

K" - K' = - μmgd

d = -(K" - K')/μmg

Substituting the values of the variables, we have

d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)

d = -(- 186.2 J)/(0.98 kg m/s²)

d = 190 m

d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m

So, a = (v² - u²)/2d

substituting the values of the variables, we have

a = (0² - (8.63 m/s)²)/(2 × 190 m)

a = -74.4769 m²/s²/380 m

a = -0.2 m/s²

Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.

t = (v - u)/a

t =(0 m/s - 8.63 m/s)/-0.2 m/s²

t = - 8.63 m/s/-0.2 m/s²

t = 43.2 s

e. The work done by friction W = fd where

= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m

W = 0.02 × 5 kg × 9.8 m/s² × 190 m

W = 186.2 J

5 0

3 years ago

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 1783

denis23 [38]

Answer:I=0.416kgm^2

Explanation:

Torque is the rotational equivalence of Force and mathematicallly, it is given by

T = r X F

Where

T=Torque

r= perpendicular distance between the applied force and the axis

F =force

T= 0.028 X 1783 =49.924 Nm

T=αI

Where α=angular acceleration produ ed

I=moment of inertia

I= T/α

I= 49.924/120

I= 0.416kgm^2

6 0

4 years ago

A 58.4-kg person, running horizontally with a velocity of +3.97 m/s, jumps onto a 14.8-kg sled that is initially at rest. (a) Ig

AURORKA [14]

Answer:0.017

Explanation:

Given

mass of person $m_1=58.4 kg$

mass of sled $m_2=14.8 kg$

velocity of person $v_1=3.97 m/s$

let u be the combined velocity of person and sled

conserving momentum

$m_1v_1=(m_1+m_2)u$

$u=\frac{m_1}{m_1+m_2}\times v_1$

$u=\frac{58.4}{58.4+14.8}\times 3.97$

$u=3.16 m/s$

(b)Total distance traveled by sled and man is 30 m

deceleration Provided by surface is $\mu _k\cdot g$

where $\mu _k$ is coefficient of kinetic friction

using $v^2-u^2=2as$

$0-3.16^2=2(-\mu _k\cdot 9.8)\cdot 30$

$\mu _k=0.0169$

6 0

4 years ago

Suppose you have an object tied to a rope and are rotating it over your head in uniform circular motion. If you increase the len

emmainna [20.7K]

Answer:

Centripetal Acceleration

In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.

Figure shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δs becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means “center seeking.”

Explanation:

8 0

3 years ago

A disk-shaped grindstone of mass 1.7 kg and radius 8 cm is spinning at 730 rev/min. After the power is shut off, a woman continu

Anastasy [175]

Answer:

0.186 N-m

Explanation:

mass of the grindstone, $m=1.7 kg$