This is my exam question be serious

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.

4 0

2 years ago

Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo

kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

$W=F*d*cos(\theta)$

here we have the work done by Paul and the friction force, so:

$W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J$

$W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J$

Now the change of energy is:

$W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s$

4 0

3 years ago

Read 2 more answers

Two charged point particle are located at two vertices of an equilateral triangle and the electric field is zero at the third ve

Debora [2.8K]

Answer:

Option E

Explanation:

In the presence of two point charges at the two vertices of an equilateral triangle, the resultant electric field at the third vertex due to these charges can not be zero whether the charges are identical or not.

The reason being that only of the x or y component of the field can be cancelled out in either case still the total field can't be reduced to zero.

This can only be achieved if another charge is present.

4 0

3 years ago

Does anyone know this?!

Valentin [98]

Answer:

2 is the numerical answer.

Explanation:

Hello there!

In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:

$KE=\frac{3}{2} \frac{R}{N_A} T$