Ask question

Ask question

All categories

2 years ago

7

in a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

ay. If the 3.5 kg book is pushed from rest through a distance of 0.96 m by the horizontal 22 N force from the broom and then has a speed of 1.36 m/s, what is the coefficient of kinetic friction between the book and floor?

1 answer:

madam [21]2 years ago

3 0

Answer:

μ=0.151

Explanation:

Given that

m= 3.5 Kg

d= 0.96 m

F= 22 N

v= 1.36 m/s

Lets take coefficient of kinetic friction = μ

Friction force Fr=μ m g

Lets take acceleration of block is a m/s²

F- Fr = m a

22 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)

a= 6.28 - 35μ m/s²

The final speed of the block is v

v= 1.36 m/s

We know that

v²= u²+ 2 a d

u= 0 m/s given that

1.36² = 2 x a x 0.96

a= 0.963 m/s²

a= 6.28 - 35μ m/s²

6.28 - 35μ = 0.963

μ=0.151

You might be interested in

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

2 years ago

If a force of 50 Newton’s was applied to an object with a mass of 500 grams, what will the objects acceleration be?

lapo4ka [179]

Answer: 100 m/s^2

F=ma

Explanation:

50N = 50 kg*m/s^2

500g = 0.5 kg

F=ma

a = F/m

a = (50 kg*m/s^2)/(0.5 kg)

a = 100 m/s^2

5 0

2 years ago

1. Two forces act on a box as follows: F= 100 N at 02 = 170. and F2= 75 N at

Setler [38]

Answer:

yfv.

Explanation:

7 0

2 years ago

Please help I’m stuck

Gnoma [55]

The weight of the box is <em>w</em> = <em>mg</em>, where <em>m</em> is the mass. So

<em>m</em> = <em>w</em>/<em>g</em> = (3893.40 N) / (9.80 m/s²) ≈ 397 kg

Then the box has density

(397 kg)/(4.60 m³) ≈ 86.4 kg/m³

which is less than the density of the given liquid, so the box will float.

8 0

2 years ago

A car travels at a speed of 40 m/s for 29.0 s;<br>what is the distance traveled by the car?

lianna [129]

Answer: 1160 m

Explanation:

Speed = distance / time. Plug in 40 m/s for speed and 29 s for time in order to get the distance, 1160 m.

8 0

1 year ago