Ask question

Ask question

All categories

snow_tiger [21]

3 years ago

14

An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov

e the horizontal. The block is displaced by 19.9 m, and the coefficient of kinetic friction is 0.209.
a. Find the work done by the 156 N force. The acceleration of gravity is 9.8 m/s^2
b. Find the magnitude of the work done by the orce of friction
c. What is the sign of the work done by the frictional force?
d. Find the work done by the normal force

1 answer:

Sav [38]3 years ago

4 0

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c) The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .

The formula for calculate the work is :

W = F*d*cosα

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α :angle that form the force (F) and displacement (d)

Known data

m = 18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.

W: Weight of the cart : In vertical direction downaward

N : Normal force : In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight of the block

W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n

Calculated of the Normal force

Newton's second law for the block in y direction :

∑Fy = m*ay ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J :magnitude

c) The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

You might be interested in

How much heat is required to raise the temperature of 0.25 kg of water from 20°C to 30°C

pentagon [3]

Answer:

10500 J/kg/*C

Explanation:

Quantity of heat required=mass of substance x specific heat capacity x change in temperature

Quantity of heat required=0.25 x 4200 x [30-20]

Quantity of heat required=0.25 x 4200 x 10

Quantity of heat required=10500 J/kg/*C

3 0

3 years ago

Which statement is true about the formation of bounds

MariettaO [177]

The formation of bonds releases energy.

4 0

3 years ago

In a certain physics experiment, carts run along a frictionless, level track. Cart A is 1 kg and Cart B is 2 kg. Initially cart

timofeeve [1]

Answer: $v=-\frac{11}{3} m/s \left$

Explanation:

Given

Mass of car A $\left ( m_a\right )=1 kg$

Mass of car B $\left ( m_b\right )=2 kg$

velocity of cart A=3 m/s

velocity of cart B=-7 m/s

Conserving momentum

$m_au_a+m_bu_b=\left ( m_a+m_b\right )v$

$v=\frac{m_au_a+m_bu_b}{\left ( m_a+m_b\right )}$

$v=-\frac{11}{3} m/s \left ( i.e. in\ left\right )$

6 0

3 years ago

Curious George is whirling a 2.0 kg bunch of bananas on a smooth floor in a circular path having a radius of 0.50 m. What force

Sedaia [141]

Answer:

The force will be $F_{c}=2.47 N$

Explanation:

Let's use the centripetal force equation.

$F_{c}=m\omega ^{2}R$

Where:

m is the mass of the bunch of bananas

ω is the angular speed

R is the radius

Now, 1 rev every 4 seconds or 0.25 rev/sec is the angular speed, but we need to write this speed in rad per second.

$\omega =0.25\frac{rev}{s}=0.25*2\pi \frac{rad}{s}=1.57 \frac{rad}{s}$

FInally, the force will be:

$F_{c}=2.0*1.57^{2}*0.5$

$F_{c}=2.0*1.57^{2}*0.5$

$F_{c}=2.47 N$

I hope it helps you!

4 0

3 years ago

A motorboat travels 92 km in 2 hours going upstream. It travels 132 km going downstream in the same amount of time. What is the

taurus [48]

Answer:

The speed on boat in still water is $56 \frac{km}{h}$ and the rate of the current is $10 \frac{km}{h}$

Explanation:

Since speed , $v= \frac{Distance\, traveled(D)}{Time\, taken(t)}$

Therefore speed of motor boat while traveling upstream is

$v_{upstream}=\frac{92}{2}\frac{km}{h}=46\frac{km}{h}$

and speed of motor boat while traveling downstream is

$v_{downstream}=\frac{132}{2}\frac{km}{h}=66\frac{km}{h}$

Let speed of boat in still water be $v_b$ and rate of current be $v_w$

Therefore $v_{upstream}=v_b-v_w=46\frac{km}{h}$ ----(A)

and $v_{downstream}=v_b+v_w=66\frac{km}{h}$ ------(B)

Adding equation (A) and (B) we get

$2v_b= (46+66) \frac{km}{h}=112 \frac{km}{h}$

=> $v_b= 56 \frac{km}{h}$ ------(C)

Substituting the value of $v_b$ in equation (A) we get

$v_w= 10 \frac{km}{h}$

Thus the speed on boat in still water is $56 \frac{km}{h}$ and the rate of the current is $10 \frac{km}{h}$

5 0

4 years ago