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kati45 [8]
3 years ago
9

sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coefficient

of kinetic friction between sledge and snow is 0.05. Howmuch work is done by the dog team pulling the sledge 1000 m at constant speed?
Physics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force
Tamiku [17]

Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

     Subsequently,

                                                  F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

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bagirrra123 [75]

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, \omega_i = 10 rad/s

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time of motion, t = 10 s

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3 years ago
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