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

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.