Question #111809

A block is being dragged along a horizontal surface by a constant horizontal force of size 45 N. It. covers 8 m in the first 2s and 8.5 in the next 1s. Find the mass of the block.

Thank you in advance!!

Thank you in advance!!

Expert's answer

Solution: To determine mass of the block we can use second Newton' law "\\vec F=m\\vec a" . The force and acceleration according the problem is directed along a horizontal surface, and we can omit the vector sign in Newton's law. The force we know "F=45N", thus we should deduce the acceleration. The problem does not specify the initial speed at which time began to count, so for the first time interval, we may write the kinematics equation in the form

(1) "S_1=v_1\\cdot t_1+a\\frac {t_1^2}{2}" , where "S_1=8m, t_1=2s" , other quantities we don't know. The similar equation we can write for next time interval

(2) "S_2=v_2\\cdot t_2+ a\\frac{t_2^2}{2}" . where "S_2=8.5m, t_2=1s"

Note that during the first time interval, the speed of the block increased in accordance with the law of equidistant motion and it became the initial speed of the second interval, i.e.

(3) "v_2=v_1+a\\cdot t_1" Substitute (3) to (2) we get

(4) "S_2=(v_1+a\\cdot t_1)\\cdot t_2+ a\\frac{t_2^2}{2}=v_1\\cdot t_2+a\\cdot t_1\\cdot t_2+a\\frac{t_2^2}{2}"

From equation (1) and (4) we can exclude unknown quantity "v_1" , then remain only one unknown "a". For determine "a" we dived (1) by "t_1" , (4) by "t_2" to find the average speed at time intervals and subtract (1) from (4).

(5) "\\frac {S_2}{t_2}-\\frac {S_1}{t_1}=v_1+a\\cdot t_1 +a\\frac {t_2}{2}-(v_1+a\\frac{t_1}{2})=a\\frac{t_1+t_2}{2}-" For acceleration we get

(6) "a=2\\cdot ( {\\frac{S_2}{t_2}-\\frac{S_1}{t_1})\/(t_1+t_2)}=2\\cdot \\frac{(8.5m\/s-4m\/s)}{3s}=3ms^{-2}"

For mass from second Newton's law we get

(7) "m=\\frac{F}{a}=\\frac{45N}{3ms^{-2}}=15kg"

Answer: The mass of the block is "15 kg"

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## Comments

Adele Scheepers09.06.22, 17:09Thanks. This was well set out and helped GREAT DEAL.

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