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2 years ago

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A climber of mass m slides down a vertical rope with an average acceleration a. What is the average frictional force exerted by

the rope on the climber?

1 answer:

Ganezh [65]2 years ago

7 0

Answer:

m(g - a)

Explanation:

Since he slides with a mass 'm' and acceleration of "a", it means he is sliding in the position of gravity. Thus, Force = ma

Now, the force due to gravity = mg

The overall acceleration of the climber is downward along the rope. So "ma" is negative. Now, the only external forces acting on the person will be the force of gravity acting downward (W = mg) and the supporting Frictional Force F_f.

Thus, the net average frictional force of exerted by the rope on the climber will be;

mg - F_f - ma = 0

F_f = mg - ma

F_f = m(g - a)

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A cylindrical tank of methanol has a mass of 40 kgand a volume of 51 L. Determine the methanol’s weight, density,and specific gr

mezya [45]

Answer:

Weight W = 392.4 N

Density $\rho$ = 784.31 $\frac{kg}{m^{3} }$

Specific gravity S = 0.78431

Force required F = 10 N

Explanation:

Given data

Mass (m) = 40 kg

Volume (V) = 0.051 $m^{3}$

Weight W = m × g

⇒ W = 40 × 9.81

⇒ W = 392.4 N

This is the weight of the methanol.

Density $\rho$ = $\frac{mass }{volume}$

⇒ $\rho$ = $\frac{40}{0.051}$

⇒ $\rho$ = 784.31 $\frac{kg}{m^{3} }$

This is the density of the methanol.

Specific gravity (S) = $\frac{\rho}{\rho_{water} }$

⇒ $S = \frac{784.31}{1000}$

⇒ S = 0.78431

This is the specific gravity of the methanol.

Force needed to accelerate this tank F = ma

⇒ F = 40 × 0.25

⇒ F = 10 N

This is the force required to accelerate the tank.

4 0

2 years ago

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w

OLga [1]

Answer:

a). va=17.23 $\frac{m}{s}$ or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

$m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65$

The motion is an inelastic collision so

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

The force of the motion is contrarest by the force of friction so

$F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}$

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

$x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s$

So the velocity final can be find using this time

$v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}$

a).

Replacing in the first equation the final velocity can find the initial velocity

$m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}$

$v_{b}=0$

$v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}$

b).

$35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}$

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

$Ek=\frac{1}{2}*m*(v)^{2}\\ Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ$

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2 years ago

A 23.5 kg object travels a distance of 85 m in 30 s. Find the momentum. (Hint: you must find speed first)

kherson [118]

Speed = 85/30 m/s
Momentum = 23.5 * 85/30 = ....

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2 years ago

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

tino4ka555 [31]

$q_{c}$ = Heat released to cold reservoir

$q_{h}$ = Heat released to hot reservoir

$W_{max}$ = maximum amount of work

$t_{c}$ = temperature of cold reservoir

$t_{h}$ = temperature of hot reservoir

we know that

$\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}$

$q_{h} = (\frac{t_{h}}{t_{c}})q_{c}$ eq-1

maximum work is given as

$W_{max}$ = $q_{h}$ - $q_{c}$

using eq-1

$W_{max}$ = $(\frac{t_{h}}{t_{c}})q_{c}$ - $q_{c}$

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2 years ago

When a student stands on a rotating table,the frictional force exerted on the student by the table is?

lakkis [162]

Answer:

Less

Explanation:

because static friction is more than rolling friction

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2 years ago