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2 years ago

Sliding friction is _ than the static friction.

Physics

1 answer:

Alona [7]2 years ago

4 0

Answer:

less

Explanation:

Sliding friction is always less than static friction. This is because in sliding friction, the bodies slide with each other and thus the effect of friction is not more. However, it does not happen in the case of static friction.

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A young kid of mass m = 36 kg is swinging on a swing. The length from the top of the swing set to the seat is L = 3.5 m. The boy

lara31 [8.8K]

Answer:

Explanation:

Given

mass of boy=36 kg

length of swing=3.5 m

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$

where v=velocity needed to complete circular path

Th-resold velocity is given by $mg-0=\frac{mv^2}{r}$

$v=\sqrt{gr}=\sqrt{9.8\times 3.5}=5.85 m/s$

So apparent weight of boy will be zero at top when it travels with a velocity of $v=\sqrt{gr}$

To get the velocity at bottom conserve energy at Top and bottom

At top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at Bottom $E_b=\frac{mv_0^2}{2}$

Comparing two as energy is conserved

$v_0^2=4gl+gl$

$v_0^2=5gL$

$v_0=\sqrt{5gL}=13.09 m/s$

Apparent weight at bottom is given by

$W=\frac{mv_0^2}{L}-mg=\frac{36\times 13.09^2}{3.5}+36\times 9.8=2115.23 N$

6 0

2 years ago

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.

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