Question

A bobsled team pushes a 132-kilogram bobsled. If the combined push of the team is 450.0 Newtons and the bobsled also experiences a force of friction of 35 Newtons, what is the acceleration of the bobsled? A. The bobsled's acceleration is 3.14 m/s2 . B. The bobsled's acceleration is 3.18 m/s2 . C. The bobsled's acceleration is 3.6 m/s2 . D. The bobsled's acceleration is 9.08 m/s2 .

Answer 1

Answer:

A. The bobsled acceleration is 3.14 m/s²

Explanation:

Given:

Mass of the bobsled (m) = 132 kg

Force of push (F) = 450.0 N

Force of friction (f) = 35 N

Let the acceleration of the bobsled be 'a' m/s².

Now, as per Newton's second law, the net force acting on a body is equal to the product of mass and acceleration.

Net force acting on the bobsled is equal to the difference of applied force and friction and is given as:

$F_{net}=F-f\\\\F_{net}=450.0\ N-35\ N=415\ N$

Now, from Newton's second law,

$F_{net}=ma\\\\a=\dfrac{F_{net}}{m}$

Plug in all the values given and solve for 'a'. This gives,

$a=\frac{415\ N}{132\ kg}\\\\a=3.14\ m/s^2$

Therefore, the acceleration is 3.14 m/s². So, option (A) is correct.