WAEC Further Mathematics Past Questions & Answers - Page 113

561.

Forces of magnitude 8N and 5N act on a body as shown above. Calculate, correct to 2 dp, the angle that the resultant makes with the horizontal.

A.

80.76°

B.

75.00°

C.

71.99°

D.

15.00°

Correct answer is A

No explanation has been provided for this answer.

562.

Forces of magnitude 8N and 5N act on a body as shown above. Calculate, correct to 2 d.p., the resultant force acting at O.

A.

14.06N

B.

13.00N

C.

9.83N

D.

8.26N

Correct answer is D

No explanation has been provided for this answer.

563.

Forces \(F_{1} = (8N, 030°)\) and \(F_{2} = (10N, 150°)\) act on a particle. Find the horizontal component of the resultant force.

A.

1.7N

B.

4.5N

C.

9.0N

D.

13.0N

Correct answer is A

\(F = F \cos \theta + F \sin \theta\)

where \(F \cos \theta = \text{horizontal component}\)

\(F \sin \theta = \text{vertical component}\)

Horizontal component of resultant = sum of horizontal compoents of individual forces

= \(8 \cos 30 + 10 \cos 150 = 6.928 - 8.66 \approxeq - 1.7N\)

564.

The sum, \(S_{n}\),  of a sequence is given by \(S_{n} = 2n^{2} - 5\). Find the 6th term

A.

112

B.

67

C.

45

D.

22

Correct answer is D

\(S_{n} = 2n^{2} - 5\)

\(T_{n} = S_{n} - S_{n - 1}\)

\(T_{6} = S_{6} - S_{5}\)

= \((2(6^{2} - 5) - (2(5^{2} - 5) = 62 - 40 = 22\)

565.

A 24N force acts on a body such that it changes its velocity from 5m/s to 9m/s in 2 secs.If the body is travelling in a straight line, calculate the distance covered in the period.

A.

22m

B.

18m

C.

14m

D.

10m

Correct answer is C

\(a = \frac{v - u}{t} = \frac{9 - 5}{2} = 2 ms^{-2}\)

\(s = ut + \frac{1}{2}at^{2} = 5(2) + \frac{1}{2}(2 \times 2^{2})\)

= \(14m\)