A 5.9 kg penguin is sliding on a frictionless horizontal surface with an initial speed of 8.0 m/s. A varying magnitude force is applied in the direction of the motion. The force starts at 19.3 N and decreases at a constant rate so that the force is 5.8 N after 10 seconds.
What is the penguin's speed after 6.1 seconds?

Answer :

Answer:

The penguin is being applied a varying force. This means that by the Newton's second law, the penguin has a varying acceleration.

[tex]a(t) =\frac{F(t)}{m}[/tex]

The force is decreased [tex](19.3 - 5.8)[/tex] 13.5 N in 10 sec. The initial value is 19.3 N. From these information the applied force as a function of time can be written as

[tex]F(t) = 19.3 - 1.35t[/tex]

(You can check that at t = 0, F = 19.3N and at t = 10s. F = 5.8N.)

So the acceleration of the penguin as a function of time can be written as

[tex]a(t) = \frac{(19.3 - 1.35t)}{m}[/tex]

Now that we know the acceleration function, we can calculate the velocity of the penguin at a given time using the following formula

[tex]v(t) = v_{0} + \int\limits^t_0 {a(t)} \, dt[/tex]

[tex]v(t=6.1) = 8 + \int\limits^{6.1}_0 {\frac{ (19.3 - 1.35t)}{5.9}} \, dt[/tex]

[tex]v(t=6.1) = 8 + \frac{1}{5.9}[ 19.3t - \frac{1.35t^2}{2}\left ] {{t=6.1} \atop {t=0}} \right.[/tex]

[tex]v(t = 6.1) = 8 + \frac{1}{5.9}[117.73 - 25.116][/tex]

[tex]v(t = 6.1) = 23.697 m/s[/tex]

Explanation:

The key factor to solve these kind of questions is to evaluate which information is given in the question and make sure that all the information is used throughout the answer. Here, no information regarding the distance is given. We need to figure out that we should use a relation between force and velocity without distance. The only thing that comes to mind is the following

[tex]v(t) = v_{0} + \int\limits^t_0 {a(t)} \, dt[/tex]

and a(t) can be found via Newton's second law:

[tex]F = ma[/tex]

The rest is just mathematical evaluation.

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