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 2 s and 8.5 m in the next 1 s. Find the mass of the block. 15kg

Can anyone please explain this sum with proper working? ​

Solution: To determine mass of the block we can use second Newton' law \vec F=m\vec a

F

=m

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=45NF=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}S

1

=v

1

⋅t

1

+a

2

t

1

2

, where S_1=8m, t_1=2s 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}S

2

=v

2

⋅t

2

+a

2

t

2

2

. where S_2=8.5m, t_2=1s 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_1v

2

=v

1

+a⋅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}S

2

=(v

1

+a⋅t

1

)⋅t

2

+a

2

t

2

2

=v

1

⋅t

2

+a⋅t

1

⋅t

2

+a

2

t

2

2

From equation (1) and (4) we can exclude unknown quantity v_1v

1

, then remain only one unknown aa. For determine aa we dived (1) by t_1t

1

, (4) by t_2t

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}-

t

2

S

2

−

t

1

S

1

=v

1

+a⋅t

1

+a

2

t

2

−(v

1

+a

2

t

1

)=a

2

t

1

+t

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}a=2⋅(

t

2

S

2

−

t

1

S

1

)/(t

1

+t

2

)=2⋅

3s

(8.5m/s−4m/s)

=3ms

−2

For mass from second Newton's law we get

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

a

F

=

3ms

−2

45N

=15kg

The mass of the block is 15 kg

college

b: |12 – x| ≤ 0.25

step-by-step explanation:      

let x be the possible lengths in inches, of the ribbons jaime cut.

we have been given that jaime is cutting strips of ribbon for a project. he tries to cut each piece to be 12 inches long, but sometimes he cuts it too long or too short by as much as 0.25 inch.  

we can represent this information in an inequality as:

by the definition of absolute value function we will get,

upon combining our inequalities we will get,

therefore, the inequality   describes all the possible lengths (x), in inches, of the ribbons jaime cut.

step-by-step explanation: