5 Physic Questions!A 4.18 kg shopping cart is accelerated by a net force of 17.70 N.What is the acceleration of the object? What is the mass of a person if they have a weight of 999.0 N near...

5 Physic Questions!

A 4.18 kg shopping cart is accelerated by a net force of 17.70 N.
What is the acceleration of the object?

What is the mass of a person if they have a weight of 999.0 N near the
earth’s surface?

A 15.8 kg object is pulled along a horizontal surface. If the coefficient
of friction between the surfaces is 0.80, what is the force of friction?

A student applies a force of 146 N on a heavy box by using a rope
held at an angle of 25 degree with the horizontal. What are the vertical
and horizontal components of the 146 N forces?

A 4.7 kg object is pulled along a horizontal surface by a force of 34.0
N. If the acceleration of the object is 1.58 m/s2, what is the coefficient
of friction between the surfaces?

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

 You need to use the second Newton's law to find the acceleration of the object such that:

F = m*a => a = F/m

a = 17.7/4.18 => a = 4.23 m/s^2

Hence,evaluating the acceleration of shopping cart yields a = 4.23 m/s^2.

2) You need to evaluate the mass of a person near Earth's surface such that:

G = m*g => 999 = m*9.8 (g = 9.8 m/s^2)

m = 999/9.8 => m = 101.93 Kg

Hence,evaluating the mass of a person near Earth's surface yields m = 101.93 Kg.

3) You need to remember the formula of friction force such that:

F_f = m*F_n

You need to find the normal force such that:

F_n - G = 0 (the object does not move vertically)

F_n = G => F_n = m*g => F_n = 15.8*9.8 = 154.84 N

F_f = 0.8*154.84 => F_f = 123.872N

Hence,evaluating the friction force under given conditions yields F_f = 123.872N.

4) You need to use decomposition of vectors to find the components of 146N force along x and y axis such that:

F_x = F*cos 25^o => F_x = 146*0.906 = 132.320 N

F_y = F*sin 25^o => F_y = 146*0.422 = 61.702N

Hence,evaluating the components of 146N force along x and y axis yields F_x =  132.320 N and  F_y =  61.702N.

5) You need to use friction force formula such that:

F_f = mu*F_n

F_n = 4.7*9.8 = 46.06N

You need to use second Newton's law to find friction force such that:

F - F_f = m*a

34 - F_f = 4.7*1.58 => F_f = 34 - 7.426

F_f = 26.574N

You need to substitute 26.574N for F_f in equation of friction force such that:

26.574N = mu*46.06N => mu = 26.547/46.06

mu = 0.576

Hence, evaluating the coefficient of friction yields mu = 0.576.

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question